Sections 12.1 – 12.5
Conic Sections, Distance Tools, and Circles
What Conic Sections Are · Distance and Midpoint · Circles · Completing the Square
Section 12.1
What Conic Sections Are
Imagine a perfectly symmetrical double cone — two cone-shaped surfaces meeting at a central point, their axes aligned vertically. If you slice through this double cone with a flat plane at different angles, the outline of the intersection traces out a curve. Depending on the angle and position of the cut, you get one of four curves.
| Cutting Plane Orientation | Resulting Curve | Core Distance Condition |
|---|
| Perpendicular to the axis | Circle | All points equidistant from one center point |
| Tilted, cuts one cone only (not parallel to edge) | Ellipse | Sum of distances to two foci is constant |
| Parallel to exactly one slant edge | Parabola | Equidistant from one focus and one directrix line |
| Steeply tilted, cuts both cones | Hyperbola | Absolute difference of distances to two foci is constant |
Diagram Instruction
Draw a double cone (two cones joined at a central point, axis vertical). Then draw four separate cutting planes and label the resulting intersection curves: horizontal cut → circle; tilted cut → ellipse; cut parallel to one slant edge → parabola; steeply tilted cut through both cones → hyperbola.
These four curves are the conic sections. They are called conic sections because they are sections — slices — of a cone. The ancient Greek mathematician Apollonius of Perga studied them systematically around 200 BCE. Today they appear throughout science, engineering, and mathematics.
In this chapter, you will study conic sections through their algebraic equations in the coordinate plane. Each conic satisfies a distance condition that translates directly into an equation in $x$ and $y$. By writing the equation in a carefully organized standard form, you can read off the key features immediately.
Section 12.2
Distance and Midpoint Review
Before working with conic sections, you need two foundational tools from coordinate geometry. Both grow naturally from the Pythagorean theorem.
Distance Formula
The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Why it works: Form a right triangle. The horizontal leg has length $|x_2 - x_1|$, the vertical leg has length $|y_2 - y_1|$. The Pythagorean theorem gives $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$. Take the positive square root.
Midpoint Formula
The midpoint of the segment from $(x_1, y_1)$ to $(x_2, y_2)$ is:
$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$
Why it works: The midpoint is in the middle. Its $x$-coordinate is the average of the two $x$-coordinates; its $y$-coordinate is the average of the two $y$-coordinates.
Example 12.2.1
Finding the distance between two points
- ProblemFind the distance between $A = (1, 3)$ and $B = (7, 11)$.
- ThinkApply the distance formula with $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (7, 11)$.
- Solve$$d = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
- CheckA 6-8-10 right triangle is a scaled 3-4-5 triangle. ✓
Answer: The distance between $A$ and $B$ is 10 units.
Example 12.2.2
Finding the midpoint of a segment
- ProblemFind the midpoint of the segment connecting $P = (-3, 4)$ and $Q = (9, -2)$.
- ThinkAverage the $x$-coordinates and $y$-coordinates separately.
- Solve$$M = \left(\frac{-3 + 9}{2},\ \frac{4 + (-2)}{2}\right) = \left(\frac{6}{2},\ \frac{2}{2}\right) = (3, 1)$$
- CheckDistance from $M=(3,1)$ to $P=(-3,4)$: $\sqrt{36+9} = \sqrt{45}$. Distance from $M$ to $Q=(9,-2)$: $\sqrt{36+9} = \sqrt{45}$. Equal. ✓
Answer: The midpoint is $(3, 1)$.
Student Tip
In conic section problems, you will frequently need the distance formula to verify a distance condition or to find the value of $a$, $b$, or $c$ in an equation. Keep the formula visible and practice it until it is automatic.
- Find the distance between $(0, 0)$ and $(5, 12)$.
- Find the distance between $(-2, 6)$ and $(4, -2)$.
- Find the midpoint of the segment from $(0, 8)$ to $(6, 2)$.
- Find the midpoint of the segment from $(-5, -3)$ to $(3, 9)$.
- The midpoint of a segment is $(4, -1)$. One endpoint is $(1, 3)$. Find the other endpoint.
- Two towns are at $(2, 5)$ and $(14, 10)$ on a coordinate grid. Find the distance between them and the location of a midpoint rest stop.
Section 12.3
Circles
A circle is the set of all points in a plane that are the same distance from a fixed point. The fixed point is called the center, and the fixed distance is called the radius. Every point on the circle is exactly $r$ units from the center — no closer, no farther.
Building the equation. Let the center be at $(h, k)$ and the radius be $r$. A point $(x, y)$ lies on the circle when its distance from $(h, k)$ equals $r$. Using the distance formula and squaring both sides:
Standard Form of a Circle
$$(x - h)^2 + (y - k)^2 = r^2$$
Center: $(h, k)$ Radius: $r$
Student Warning — Sign of the Center
The equation uses subtraction: $(x - h)$ and $(y - k)$. If the equation reads $(x + 3)^2 + (y - 2)^2 = 25$, rewrite the first factor as $(x - (-3))^2$, so the center is $(-3, 2)$, not $(3, 2)$. Always read the center as the values that make each factor equal to zero.
Example 12.3.1
Writing the equation of a circle
- ProblemWrite the equation of the circle with center $(4, -1)$ and radius 7.
- ThinkSubstitute $h = 4$, $k = -1$, and $r = 7$ into the standard form.
- Solve$$(x - 4)^2 + (y - (-1))^2 = 7^2 \implies (x - 4)^2 + (y + 1)^2 = 49$$
- CheckTest $(11, -1)$, which is 7 units right of center: $(11-4)^2 + (-1+1)^2 = 49 + 0 = 49$. ✓
Answer: $(x - 4)^2 + (y + 1)^2 = 49$
Example 12.3.2
Identifying center and radius
- ProblemIdentify the center and radius of $(x + 2)^2 + (y - 5)^2 = 36$.
- ThinkRewrite $(x + 2)$ as $(x - (-2))$. So $h = -2$, $k = 5$, and $r^2 = 36$, giving $r = 6$.
Answer: Center $(-2, 5)$, radius $6$.
Diagram Instruction — Graphing a Circle
Plot the center at $(4, -1)$. From the center, mark four points by moving $r = 7$ units in each cardinal direction: right to $(11, -1)$, left to $(-3, -1)$, up to $(4, 6)$, down to $(4, -8)$. Sketch the circle passing through all four points with smooth, symmetric curves.
Section 12.4
Equations of Circles
When you expand the standard form of a circle, you get the general form: $x^2 + y^2 + Dx + Ey + F = 0$. The center and radius are not visible in general form. The standard form is more useful for graphing and interpreting the circle.
Example 12.4.1
Writing standard form from center and a point
- ProblemWrite the equation of the circle with center $(3, -2)$ that passes through $(7, 1)$.
- ThinkThe radius is the distance from center to the given point.
- Solve$$r = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
$$(x - 3)^2 + (y + 2)^2 = 25$$
- CheckSubstitute $(7, 1)$: $(7-3)^2 + (1+2)^2 = 16 + 9 = 25$. ✓
Answer: $(x - 3)^2 + (y + 2)^2 = 25$
Example 12.4.2
Expanding to general form
- ProblemExpand $(x - 3)^2 + (y + 2)^2 = 25$ into general form.
- Solve$$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 25$$
$$x^2 + y^2 - 6x + 4y + 13 = 25$$
$$x^2 + y^2 - 6x + 4y - 12 = 0$$
Answer: $x^2 + y^2 - 6x + 4y - 12 = 0$
Section 12.5
Completing the Square for Circles
When a circle equation is given in general form, you must rewrite it in standard form before you can identify the center and radius or graph it. The technique is completing the square, applied separately to the $x$-terms and the $y$-terms.
To complete the square for $x^2 + bx$: take half the coefficient of $x$, square it, and add it. This creates a perfect square trinomial: $x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$. Whatever you add to one side of an equation, you must add to the other.
Example 12.5.1
Converting general form to standard form
- ProblemRewrite $x^2 + y^2 - 6x + 4y - 12 = 0$ in standard form. Identify the center and radius.
- ThinkGroup $x$-terms and $y$-terms, move the constant to the right, then complete the square for each group.
- Solve$$(x^2 - 6x) + (y^2 + 4y) = 12$$
\text{For } x: \text{ half of } {-6} \text{ is } {-3}, \text{ squared is } 9. \text{ Add 9 to both sides.}
\text{For } y: \text{ half of } 4 \text{ is } 2, \text{ squared is } 4. \text{ Add 4 to both sides.}
$$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$$
$$(x - 3)^2 + (y + 2)^2 = 25$$
Answer: Center $(3, -2)$, radius $5$.
Example 12.5.2
Completing the square with a leading coefficient
- ProblemRewrite $4x^2 + 4y^2 - 8x + 24y + 4 = 0$ in standard form.
- ThinkThe leading coefficient is 4. Divide every term by 4 first, then complete the square.
- Solve$$x^2 + y^2 - 2x + 6y + 1 = 0$$
$$(x^2 - 2x) + (y^2 + 6y) = -1$$
\text{Complete for } x: \text{ add } 1. \quad \text{Complete for } y: \text{ add } 9.
$$(x^2 - 2x + 1) + (y^2 + 6y + 9) = -1 + 1 + 9$$
$$(x - 1)^2 + (y + 3)^2 = 9$$
Answer: Center $(1, -3)$, radius $3$.
Student Warning — Divide First
Completing the square only works cleanly when the quadratic terms have coefficient 1. If the equation has $4x^2$ and $4y^2$, divide the entire equation by 4 before proceeding. Forgetting this step produces incorrect centers and radii.
- Write the equation of the circle with center $(-3, 7)$ and radius $5$.
- Write the equation of the circle with center $(0, -4)$ and radius $\sqrt{11}$.
- Identify center and radius: $(x - 6)^2 + (y + 1)^2 = 100$.
- Identify center and radius: $(x + 4)^2 + y^2 = 8$.
- Convert to standard form: $x^2 + y^2 - 4x + 6y - 3 = 0$.
- Convert to standard form: $x^2 + y^2 + 10x - 2y + 17 = 0$.
- Convert to standard form: $x^2 + y^2 + 8x - 10y + 5 = 0$.
- Convert to standard form: $2x^2 + 2y^2 - 12x + 4y - 6 = 0$.
For problems 5–8, identify the center and radius after converting.
- A circular fountain is designed with center at $(5, 5)$ on an architect's grid and radius 8 units. Write the equation of the fountain's boundary.
- A Ferris wheel is centered at $(0, 30)$ on a coordinate model where units are in feet, and has radius 30 feet. Write the equation. What are the lowest and highest points on the wheel?
Error Problem 1
Center of $(x+3)^2 + (y-1)^2 = 16$ is $(3, -1)$ and radius is 16.
Identify both errors. State the correct center and radius.
Error Problem 2
For $x^2 + y^2 + 6x - 4y = 7$: adds 9 and 4 to the left side only, getting $(x+3)^2 + (y-2)^2 = 7$.
Explain why the right side is wrong. What should the equation be?
Section 12.6
Parabolas as Conic Sections
Focus · Directrix · The Value p · Four Opening Directions · Applications
Section 12.6
Parabolas as Conic Sections
You know parabolas as the graphs of quadratic functions. In this section, you will see the parabola from a new geometric angle — as a conic section defined by a distance condition. This definition is more general and more powerful than the function definition, and it explains why parabolic shapes are so useful in engineering.
A parabola is the set of all points in a plane that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. For every point $P$ on the parabola, the distance from $P$ to the focus equals the distance from $P$ to the directrix (measured perpendicularly).
The point on the parabola closest to both the focus and the directrix is the vertex. The vertex lies midway between focus and directrix. The distance from the vertex to the focus is called $p$. The directrix is also $p$ units from the vertex, on the opposite side from the focus.
Standard Forms of a Parabola
Vertex at $(h, k)$, opens vertically:
$$(x - h)^2 = 4p(y - k)$$
Focus: $(h,\ k+p)$ · Directrix: $y = k - p$ · Opens up if $p > 0$, down if $p < 0$
Vertex at $(h, k)$, opens horizontally:
$$(y - k)^2 = 4p(x - h)$$
Focus: $(h+p,\ k)$ · Directrix: $x = h - p$ · Opens right if $p > 0$, left if $p < 0$
Connecting to Algebra I
In Algebra I you learned $y = ax^2$ as a parabola with vertex at the origin. This is a special case of conic form. The equation $x^2 = 4py$ corresponds to $y = \frac{1}{4p}x^2$, so $a = \frac{1}{4p}$. The focus distance $p$ and the leading coefficient $a$ are inversely related — a narrow parabola (large $|a|$) has a focus close to the vertex (small $|p|$).
Example 12.6.1
Finding the focus and directrix
- ProblemFind the focus and directrix of $x^2 = 12y$.
- ThinkThis is the form $x^2 = 4py$ with vertex at the origin. Match: $4p = 12$, so $p = 3$.
- SolveVertex: $(0, 0)$. Focus: $(0, 3)$ (above the vertex, since $p > 0$). Directrix: $y = -3$.
Answer: Focus $(0, 3)$, directrix $y = -3$.
Example 12.6.2
Writing the equation of a parabola
- ProblemA parabola has vertex $(2, -1)$ and focus $(2, 3)$. Write its equation.
- ThinkSame $x$-coordinate for vertex and focus → opens vertically. Focus is above the vertex → $p > 0$. Distance: $p = 3 - (-1) = 4$.
- Solve$$(x - 2)^2 = 4(4)(y - (-1)) \implies (x - 2)^2 = 16(y + 1)$$
- CheckAt vertex $(2, -1)$: $(2-2)^2 = 16(-1+1) \implies 0 = 0$. ✓
Answer: $(x - 2)^2 = 16(y + 1)$
Example 12.6.3
Answering the opening question — satellite dish placement
- ProblemA satellite dish is 4 meters wide and 0.5 meters deep, vertex at the origin. Find the focus location.
- ThinkThe rim passes through $(\pm 2, 0.5)$. The dish opens upward: $x^2 = 4py$. Substitute a rim point.
- Solve$$4 = 4p(0.5) = 2p \implies p = 2$$
- CheckThe focus is 2 meters above the vertex, inside the 0.5-meter-deep dish — that seems impossible until you note the dish is 4 m wide and the focus sits on the axis of symmetry, not at the rim. In this model the vertex is at the bottom center and the dish opens upward; the focus is 2 m above the vertex. ✓
Answer: The receiver (focus) should be placed 2 meters from the vertex along the axis of the dish.
MGU Library Connection
- Parabolic reflector principle — MGU Physics: Wave Optics and Electromagnetic Radiation
- Antenna and acoustic design — MGU Engineering Applications
- Solar concentrator geometry — MGU Environmental Science and Engineering
- Find the focus and directrix of $x^2 = 8y$.
- Find the focus and directrix of $y^2 = -20x$.
- Find the focus and directrix of $x^2 = -4y$.
- Write the equation: vertex at origin, focus at $(0, -5)$.
- Write the equation: vertex at $(1, 3)$, focus at $(1, 6)$.
- Write the equation: vertex at $(-2, 4)$, directrix $x = -6$.
- A road tunnel has a parabolic arch 12 m wide and 8 m tall. Using a coordinate system with the vertex at the top of the arch, write the equation.
- A solar cooker is 60 cm wide and 15 cm deep. Find the location of the focus where cooking occurs.
Sections 12.7 – 12.8
Ellipses
Geometric Definition · $a^2 = b^2 + c^2$ · Horizontal and Vertical Forms · Applications
Section 12.7
Ellipses
An ellipse is the set of all points in a plane for which the sum of the distances to two fixed points is constant. The two fixed points are called the foci (singular: focus).
Imagine placing two pushpins in a board and stretching a loose string between them. Holding a pencil against the string and keeping it taut as you trace a full curve gives you an ellipse. The two pins are the foci. No matter where the pencil is on the ellipse, the total string length from pin to pencil to pin remains constant.
Every ellipse has a major axis — the longer axis, running through both foci and connecting the two vertices — and a minor axis — the shorter axis, perpendicular to the major axis at the center, connecting the two co-vertices.
Key Measurements of an Ellipse
Let $2a$ = the constant sum of distances (total string length). Then:
$a$ = distance from center to each vertex (along the major axis)
$b$ = distance from center to each co-vertex (along the minor axis)
$c$ = distance from center to each focus
$$a^2 = b^2 + c^2 \quad \text{(equivalently, } c^2 = a^2 - b^2\text{)}$$
Note: $a > b$ (major axis is longer than minor axis) and $a > c$ (vertices are farther from center than foci).
Why $a^2 = b^2 + c^2$? Consider the point at the end of the minor axis (a co-vertex). Its distance to each focus is, by symmetry, exactly $a$ (the two distances must be equal and sum to $2a$). The center, a focus, and the co-vertex form a right triangle with legs $b$ and $c$ and hypotenuse $a$. The Pythagorean theorem gives $a^2 = b^2 + c^2$.
Diagram Instruction
Draw an ellipse centered at the origin with the major axis along the $x$-axis. Label: center at origin; foci at $(-c, 0)$ and $(c, 0)$; vertices at $(-a, 0)$ and $(a, 0)$; co-vertices at $(0, b)$ and $(0, -b)$. Draw the right triangle with legs $b$ (vertical) and $c$ (horizontal) and hypotenuse $a$, connecting the center, a focus, and a co-vertex to illustrate $a^2 = b^2 + c^2$.
Section 12.8
Equations of Ellipses
Standard Forms of an Ellipse (Center at $(h, k)$)
Horizontal major axis ($a^2$ under $x$-term, $a > b$):
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
Vertices: $(h \pm a,\ k)$ · Co-vertices: $(h,\ k \pm b)$ · Foci: $(h \pm c,\ k)$
Vertical major axis ($a^2$ under $y$-term, $a > b$):
$$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$$
Vertices: $(h,\ k \pm a)$ · Co-vertices: $(h \pm b,\ k)$ · Foci: $(h,\ k \pm c)$
Key Observation
The larger denominator ($a^2$) tells you the direction of the major axis. If $a^2$ is under the $x$-term, the major axis is horizontal. If $a^2$ is under the $y$-term, the major axis is vertical. Always compare denominators before deciding.
Example 12.8.1
Finding features of an ellipse from standard form
- ProblemFor $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$, find the center, vertices, co-vertices, and foci.
- ThinkLarger denominator is 25 under the $x$-term → major axis is horizontal. $a^2 = 25$, $a = 5$. $b^2 = 9$, $b = 3$. Center at origin.
- Solve$$c^2 = a^2 - b^2 = 25 - 9 = 16 \implies c = 4$$
Center: $(0,0)$ · Vertices: $(\pm 5, 0)$ · Co-vertices: $(0, \pm 3)$ · Foci: $(\pm 4, 0)$
- CheckFrom vertex $(5, 0)$: distance to $(4,0)$ is 1, distance to $(-4,0)$ is 9. Sum $= 10 = 2a$. ✓
Answer: Center $(0,0)$; vertices $(\pm 5, 0)$; co-vertices $(0, \pm 3)$; foci $(\pm 4, 0)$.
Example 12.8.2
Writing the equation of an ellipse from features
- ProblemWrite the equation of the ellipse with center $(3, -2)$, vertices $(3, 4)$ and $(3, -8)$, co-vertices $(6, -2)$ and $(0, -2)$.
- ThinkVertices share the same $x$-coordinate as the center → vertical major axis. Distance from center $(3,-2)$ to vertex $(3,4)$: $a = 6$. Distance from center to co-vertex $(6,-2)$: $b = 3$.
- Solve$$\frac{(x-3)^2}{9} + \frac{(y+2)^2}{36} = 1$$
Answer: $\dfrac{(x-3)^2}{9} + \dfrac{(y+2)^2}{36} = 1$
Example 12.8.3
Finding the equation from foci and a vertex
- ProblemAn ellipse has foci at $(-3, 0)$ and $(3, 0)$ and a vertex at $(5, 0)$. Write its equation.
- ThinkFoci on $x$-axis → horizontal major axis. $c = 3$, $a = 5$. Find $b^2 = a^2 - c^2$.
- Solve$$b^2 = 25 - 9 = 16 \implies \frac{x^2}{25} + \frac{y^2}{16} = 1$$
Answer: $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$
- Find all features of $\dfrac{x^2}{49} + \dfrac{y^2}{16} = 1$.
- Find all features of $\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1$.
- Find all features of $\dfrac{(x-1)^2}{36} + \dfrac{(y+3)^2}{11} = 1$.
- Write the equation: center $(0,0)$, horizontal major axis, $a = 8$, $b = 5$.
- Write the equation: vertices $(\pm 10, 0)$, foci $(\pm 6, 0)$.
- Write the equation: center $(-2, 4)$, vertical major axis, $a = 7$, $b = 3$.
- Earth's orbit has a major axis of approximately 300 million km total and the Sun is approximately 2.5 million km from the center of the orbit. Find $b$ and write an equation modeling the orbit (in millions of km, center at origin).
- A whispering gallery is an elliptical room 80 feet long (major axis) and 40 feet wide (minor axis). How far from each end of the room are the focal points where whisperers should stand?
Sections 12.9 – 12.10
Hyperbolas
Geometric Definition · $c^2 = a^2 + b^2$ · Asymptotes · Two Branches
Section 12.9
Hyperbolas
A hyperbola is the set of all points in a plane for which the absolute value of the difference of distances to two fixed points (foci) is constant. Compare this with the ellipse: an ellipse adds distances; a hyperbola subtracts them (and takes the absolute value). This one change in the distance condition produces a dramatically different curve.
A hyperbola consists of two separate branches, curving away from the center in opposite directions. Each branch curves toward one vertex. Unlike the ellipse, the foci lie outside the vertices — they are farther from the center than the nearest point of the curve.
Key Measurements of a Hyperbola
The constant difference of distances is $2a$. Then:
$a$ = distance from center to each vertex (along the transverse axis)
$b$ = half-length of the conjugate axis (used for the asymptotes)
$c$ = distance from center to each focus
$$c^2 = a^2 + b^2$$
Note: $c > a$ for a hyperbola (the foci are farther from center than the vertices). This is the opposite of the ellipse relationship.
Diagram Instruction
Draw a hyperbola with a horizontal transverse axis. Mark the center at the origin; vertices at $(-a, 0)$ and $(a, 0)$; foci at $(-c, 0)$ and $(c, 0)$. Draw both branches curving outward from the vertices. Draw dashed asymptote lines through the center. Draw the dashed central rectangle with half-width $a$ and half-height $b$ — the asymptotes pass through its corners.
Section 12.10
Equations of Hyperbolas
Standard Forms of a Hyperbola (Center at $(h, k)$)
Horizontal transverse axis (opens left and right):
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
Vertices: $(h \pm a,\ k)$ · Foci: $(h \pm c,\ k)$ · Asymptotes: $y - k = \pm\dfrac{b}{a}(x-h)$
Vertical transverse axis (opens up and down):
$$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$
Vertices: $(h,\ k \pm a)$ · Foci: $(h,\ k \pm c)$ · Asymptotes: $y - k = \pm\dfrac{a}{b}(x-h)$
Ellipse vs. Hyperbola — Quick Comparison
An ellipse has a plus sign between fractions and equals 1. The foci are inside ($c < a$), and $c^2 = a^2 - b^2$. A hyperbola has a minus sign and equals 1. The foci are outside ($c > a$), and $c^2 = a^2 + b^2$. The positive term determines which direction the hyperbola opens.
Example 12.10.1
Identifying features of a hyperbola
- ProblemFind the center, vertices, foci, and asymptotes of $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
- ThinkThe $x$-term is positive → horizontal transverse axis. $a^2 = 16$, $a = 4$. $b^2 = 9$, $b = 3$. Center at origin.
- Solve$$c^2 = 16 + 9 = 25 \implies c = 5$$
Center: $(0,0)$ · Vertices: $(\pm 4, 0)$ · Foci: $(\pm 5, 0)$ · Asymptotes: $y = \pm\dfrac{3}{4}x$
Answer: Center $(0,0)$; vertices $(\pm 4, 0)$; foci $(\pm 5, 0)$; asymptotes $y = \pm\frac{3}{4}x$.
Example 12.10.2
A hyperbola with a vertical transverse axis
- ProblemFind all features of $\dfrac{(y-1)^2}{25} - \dfrac{(x+2)^2}{11} = 1$.
- Think$y$-term is positive → vertical transverse axis. $a^2 = 25$, $a = 5$. $b^2 = 11$. Center at $(-2, 1)$.
- Solve$$c^2 = 25 + 11 = 36 \implies c = 6$$
Center: $(-2, 1)$ · Vertices: $(-2, 6)$ and $(-2, -4)$ · Foci: $(-2, 7)$ and $(-2, -5)$
Asymptotes: $y - 1 = \pm\dfrac{5}{\sqrt{11}}(x + 2)$
Answer: Center $(-2, 1)$; vertices $(-2, 6)$ and $(-2, -4)$; foci $(-2, 7)$ and $(-2, -5)$; asymptotes $y - 1 = \pm\dfrac{5}{\sqrt{11}}(x+2)$.
Example 12.10.3
Writing a hyperbola equation from features
- ProblemWrite the equation of the hyperbola with center at the origin, vertices at $(\pm 3, 0)$, and foci at $(\pm 5, 0)$.
- ThinkHorizontal transverse axis (vertices on $x$-axis). $a = 3$, $c = 5$. Find $b^2 = c^2 - a^2$.
- Solve$$b^2 = 25 - 9 = 16 \implies \frac{x^2}{9} - \frac{y^2}{16} = 1$$
Answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$
The Asymptote Rectangle
A useful graphing aid for hyperbolas is the central rectangle. Draw a rectangle centered at $(h, k)$ with horizontal half-length $a$ and vertical half-length $b$. The diagonals of this rectangle, extended as lines through the center, give the asymptotes. The branches of the hyperbola pass through the vertices and curve outward, approaching the asymptotes but never reaching them.
- Find all features of $\dfrac{x^2}{4} - \dfrac{y^2}{32} = 1$.
- Find all features of $\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1$.
- Find all features of $\dfrac{(x+3)^2}{25} - \dfrac{(y-4)^2}{144} = 1$.
- Write the equation: center $(0,0)$, horizontal transverse axis, $a = 6$, $b = 8$.
- Write the equation: vertices $(0, \pm 5)$, foci $(0, \pm 13)$.
- Write the equation: center $(-1, 3)$, vertical transverse axis, $a = 4$, $c = 5$.
Section 12.12
Conics in Astronomy, Engineering, and Design
Conic sections are not abstract curiosities. They appear repeatedly in the physical world because they are the natural outcome of inverse-square laws, reflective properties, and orbital mechanics.
Planetary Orbits — Ellipses
In 1609, Johannes Kepler published his first law of planetary motion: every planet orbits the Sun in an ellipse, with the Sun located at one focus. This discovery, later derived from Newton's law of universal gravitation, means that no planet travels in a perfect circle. Earth's orbit is nearly circular, but the orbits of Mercury, Mars, and comets can be noticeably elongated.
Eccentricity measures how elongated an ellipse is:
Eccentricity
$$e = \frac{c}{a} \quad (0 \leq e < 1 \text{ for an ellipse})$$
When $e = 0$, the ellipse is a perfect circle. As $e \to 1$, the ellipse becomes more elongated and the foci move toward the vertices.
Example 12.12.1
Earth's elliptical orbit
- ProblemEarth's orbit has semi-major axis $a \approx 149.6$ million km and eccentricity $e \approx 0.017$. Find $c$ and $b$.
- Solve$$c = ea = 0.017 \times 149.6 \approx 2.54 \text{ million km}$$
$$b = \sqrt{a^2 - c^2} = \sqrt{149.6^2 - 2.54^2} \approx 149.58 \text{ million km}$$
- InterpretEarth's orbit is very close to circular — the Sun is only about 2.5 million km off-center from an orbit roughly 150 million km in radius. The eccentricity $e \approx 0.017$ confirms this near-circularity.
Parabolic Reflectors
Parabolas have a remarkable reflective property: any ray traveling parallel to the axis of a parabola, upon hitting the curve, reflects directly toward the focus. Conversely, any ray from the focus reflects outward parallel to the axis. This property makes parabolas essential in satellite dishes, radio telescopes, reflecting telescopes, automobile headlights, and solar cookers.
Hyperbolas in Navigation and Engineering
The difference-of-distances definition of a hyperbola is the basis of LORAN (long-range navigation), a historical radio navigation system. Two fixed transmitters broadcast signals simultaneously. A receiver measures the difference in arrival times, which corresponds to a constant difference in distances — placing the receiver on a hyperbola with the transmitters as foci. Multiple signal pairs allow precise location.
Hyperbolic shapes also appear in the design of cooling towers for power plants. These towers have a hyperboloid shape — generated by rotating a hyperbola around its conjugate axis — which is structurally strong, uses less material than a cylinder, and creates natural airflow.
Whispering galleries use elliptical ceilings to concentrate sound from one focus to the other. The National Statuary Hall in the U.S. Capitol and the Mormon Tabernacle in Salt Lake City are famous examples.
Example 12.12.2
Parabolic arch bridge
- ProblemA parabolic arch bridge spans 200 feet with the top of the arch 80 feet above the base. Vertex at origin, opening downward. Write the equation and find the height 50 feet from center.
- ThinkThe rim of the arch passes through $(\pm 100, -80)$ (span 200 ft, arch drops 80 ft). Equation form: $x^2 = 4py$ with $p < 0$.
- Solve$$100^2 = 4p(-80) \implies 10000 = -320p \implies p = -31.25$$
$$x^2 = -125y \quad\text{or equivalently}\quad y = -\frac{x^2}{125}$$
\text{At } x = 50: \quad y = -\frac{2500}{125} = -20
Answer: The arch is $80 - 20 = \mathbf{60}$ feet above the ground at 50 feet from center.
MGU Library Connection
- Orbital mechanics (Kepler's laws) — MGU Physics: Classical Mechanics
- Parabolic optics (telescope design) — MGU Physics: Wave Optics
- LORAN and GPS navigation — MGU Engineering: Signal Systems
- Parametric and polar conics — MGU Precalculus, Chapter 8
- Volumes of revolution of conic shapes — MGU Calculus I, Chapter 6
- A comet has an elliptical orbit with the Sun at one focus, semi-major axis 500 million km, and eccentricity 0.9. Find $c$, $b$, and the distance from the comet to the Sun at its closest point (perihelion).
- An elliptical billiard table is 12 feet long and 8 feet wide. A pocket is placed at one focus. Find the locations of both foci.
- A flashlight reflector is parabolic with the bulb 2 cm from the vertex. If the reflector opening is 10 cm in diameter, how deep is the reflector?
- A stone arch bridge spans a river with a semi-elliptical arch 60 meters wide and 20 meters tall at center. Write the equation of the ellipse (center at origin) and find the arch height 15 meters from center.
Sections 12.13 – 12.14
Common Mistakes and the Road Ahead
Seven Recurring Errors · Error Analysis · Preparing for Precalculus
Section 12.13
Common Conic Section Mistakes
Mistake 1 — Sign Error in the Center
Center of $(x+3)^2 + (y-5)^2 = 36$ written as $(3, 5)$.
Fix: Rewrite $(x+3) = (x - (-3))$. The center uses the values that make each factor zero: $h = -3$, $k = 5$. Always read the sign actively.
Mistake 2 — Not Multiplying by the Leading Factor When Balancing
Completing the square in $4(x^2 - 6x)$: adds 9 inside and only 9 to the right side.
Fix: Adding 9 inside $4(\cdots)$ adds $4 \times 9 = 36$ to the expression. Add 36 to the right side, not just 9.
Mistake 3 — Using the Wrong $c^2$ Formula
Using $c^2 = a^2 - b^2$ for a hyperbola (the ellipse formula).
Fix: For an ellipse, $c < a$ so $c^2 = a^2 - b^2$. For a hyperbola, $c > a$ so $c^2 = a^2 + b^2$. If your $c$ comes out less than $a$ for a hyperbola, you used the wrong formula.
Mistake 4 — Misidentifying the Major Axis of an Ellipse
For $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$: major axis identified as horizontal because the "$x$-term comes first."
Fix: The major axis goes with the larger denominator. Since $25 > 9$ and 25 is under the $y$-term, the major axis is vertical.
Mistake 5 — Misidentifying the Direction a Hyperbola Opens
$\dfrac{y^2}{9} - \dfrac{x^2}{4} = 1$ identified as opening left and right.
Fix: The positive term determines the direction. The $y$-term is positive (first) → opens up and down.
Mistake 6 — Forgetting to Square the Radius
Circle with radius 7 written as $(x-2)^2 + (y+3)^2 = 7$.
Fix: Standard circle form has $r^2$ on the right. If the radius is 7, the equation uses $49$. If the equation shows 49, the radius is $\sqrt{49} = 7$, not 49.
Mistake 7 — Forgetting Extraneous Solutions
When solving a system (e.g., line intersects circle), keeping both algebraic solutions without checking them in the original equation.
Fix: Always substitute candidate solutions back into the original equation to verify. Squaring and substituting can introduce extraneous values.
For each problem, identify the error and write the correct solution.
- A student writes the center of $(x - 4)^2 + (y + 7)^2 = 81$ as $(4, 7)$ and the radius as 81.
- A student converts $4(x^2 - 4x) + y^2 = 8$ by adding 4 inside the first group and writing $4(x-2)^2 + y^2 = 12$.
- A student has a hyperbola with $a = 3$ and $b = 4$ and computes $c^2 = 9 - 16 = -7$, concluding there are no foci.
- A student reads $\dfrac{x^2}{16} + \dfrac{y^2}{36} = 1$ and says the major axis is horizontal with vertices at $(\pm 4, 0)$.
- A student rewrites $2x^2 + 2y^2 + 8x - 12y + 6 = 0$ as $(x+2)^2 + (y-3)^2 = 7$ without dividing by 2 first. What should the correct equation be?
Section 12.14
Preparing for Precalculus Geometry
You have now built the complete algebraic foundation for conic sections. You can write equations, identify features, graph, convert between forms, and apply conic sections to real-world problems. In Precalculus, you will encounter conic sections again — and go considerably further.
Parametric Equations. Conic sections can be described using parametric equations, where both $x$ and $y$ are functions of a third variable (usually $t$ or $\theta$). The unit circle, for instance, becomes $x = \cos\theta$, $y = \sin\theta$. Parametric forms are natural for describing motion along curves.
Polar Form of Conics. In polar coordinates, all four conic sections unify into a single equation involving eccentricity:
Polar Form of Conics (Precalculus Preview)
$$r = \frac{ed}{1 \pm e\cos\theta}$$
The value of $e$ determines the conic: $e < 1$ → ellipse; $e = 1$ → parabola; $e > 1$ → hyperbola. All four conics are members of one continuous family, separated only by eccentricity.
Rotated Conics. When the general form includes an $xy$-term ($Bxy$), the conic is rotated relative to the coordinate axes. Precalculus covers rotation of axes to eliminate this term and identify the resulting conic.
Calculus Applications. In Calculus, you will find areas, arc lengths, and volumes related to conic sections, and you will use parametric and polar equations to describe and analyze curved paths. The ellipse you studied algebraically here reappears as the provable consequence of Newton's law of universal gravitation.
| Conic | Algebra I Connection | Algebra II Standard Form | Precalculus Extension |
|---|
| Parabola | $y = ax^2 + bx + c$ | $(x-h)^2 = 4p(y-k)$ | Polar/parametric; conic from focus-directrix |
| Circle | Distance formula | $(x-h)^2 + (y-k)^2 = r^2$ | Parametric: $x = r\cos\theta,\ y = r\sin\theta$ |
| Ellipse | Not in Algebra I | $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ | Kepler's laws; polar conics |
| Hyperbola | Not in Algebra I | $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ | Navigation models; polar conics; vectors |
Chapter 12 — Back Matter
Chapter Summary
- Sec 12.1Conic sections (circle, parabola, ellipse, hyperbola) are the curves formed by intersecting a plane with a double cone. Each is defined by a distance condition and described by an algebraic equation in standard form.
- Sec 12.2Distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Midpoint formula: $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. Both are essential tools throughout the chapter.
- Sec 12.3 – 12.4A circle is the set of points equidistant from a center. Standard form: $(x-h)^2 + (y-k)^2 = r^2$. Center $(h, k)$, radius $r$. General form $x^2 + y^2 + Dx + Ey + F = 0$ is less useful for graphing.
- Sec 12.5Complete the square separately for $x$-terms and $y$-terms to convert general form to standard form. When a leading coefficient is not 1, divide first. Add the completed values to both sides, multiplied by any factor in front.
- Sec 12.6A parabola is equidistant from a focus and directrix. Standard forms use $4p$, where $p$ is the vertex-to-focus distance. Sign of $p$ determines opening direction. Parabolic shapes focus parallel rays at the focus.
- Sec 12.7 – 12.8An ellipse has a constant sum of distances to two foci. Larger denominator ($a^2$) identifies the major axis direction. $c^2 = a^2 - b^2$ (foci inside, $c < a$). Applications include planetary orbits and whispering galleries.
- Sec 12.9 – 12.10A hyperbola has a constant absolute difference of distances. Two branches open toward the positive term's axis. $c^2 = a^2 + b^2$ (foci outside, $c > a$). Asymptotes pass through the center; use the central rectangle to graph them.
- Sec 12.11From $Ax^2 + Cy^2 + Dx + Ey + F = 0$: $A = C$ → circle; same sign, $A \neq C$ → ellipse; opposite signs → hyperbola; one coefficient is zero → parabola. Complete the square to convert to standard form and find features.
- Sec 12.12Conics model planetary orbits (Kepler's first law, ellipses), parabolic reflectors (satellite dishes, telescopes), navigation systems (hyperbolas, LORAN), architectural arches (parabolas), and elliptical whispering galleries. Eccentricity $e = c/a$ measures elongation.
- Sec 12.13 – 12.14Common mistakes involve sign errors in centers, unbalanced completing-the-square with leading coefficients, swapped ellipse/hyperbola formulas, and misidentifying major axis direction or hyperbola opening. Precalculus extends conics to parametric, polar, and rotated forms.
Key Terms Review
conic section
A curve formed by the intersection of a plane with a double cone. Includes circles, parabolas, ellipses, and hyperbolas.
focus / foci
Fixed point(s) used in the geometric definition of parabolas, ellipses, and hyperbolas.
directrix
A fixed line used together with the focus to define a parabola.
circle
The set of all points equidistant from a fixed center. Equation: $(x-h)^2 + (y-k)^2 = r^2$.
radius
The fixed distance from the center of a circle to any point on it.
parabola
The set of all points equidistant from a focus and a directrix.
vertex (parabola)
The turning point of a parabola; the midpoint between focus and directrix.
p (parabola)
The distance from the vertex to the focus; also the distance from the vertex to the directrix.
ellipse
The set of all points for which the sum of distances to two foci is constant ($= 2a$).
major axis
The longer axis of an ellipse, running through both foci and both vertices.
minor axis
The shorter axis of an ellipse, perpendicular to the major axis at the center.
co-vertex
An endpoint of the minor axis of an ellipse.
hyperbola
The set of all points for which the absolute difference of distances to two foci is constant ($= 2a$).
transverse axis
The axis of a hyperbola connecting its two vertices; lies along the direction the hyperbola opens.
asymptote
A line that a branch of a hyperbola approaches but never reaches; each hyperbola has two.
eccentricity
$e = c/a$; measures how elongated an ellipse is. $e = 0$ is a circle; $e \to 1$ is a very elongated ellipse.
standard form
A form of the conic equation that displays center (or vertex), $a$, $b$, $r$, and $p$ directly.
general form
$Ax^2 + Cy^2 + Dx + Ey + F = 0$; the conic type is identified by comparing $A$ and $C$.
completing the square
An algebraic technique for rewriting a quadratic expression as a perfect square trinomial; used to convert general form to standard form.
Concept Review Questions
- Name the four conic sections. For each one, state the distance condition that defines it.
- What is the difference between the focus of a parabola and the foci of an ellipse?
- How does the sign of $p$ in a parabola's equation determine the direction it opens?
- How do you determine whether an ellipse has a horizontal or vertical major axis from its equation?
- State the relationship between $a$, $b$, and $c$ for an ellipse. How is it different for a hyperbola?
- Why does a hyperbola have two branches while an ellipse is one connected curve?
- What does eccentricity measure for an ellipse? What value of eccentricity corresponds to a circle?
- How can you identify which type of conic a general equation represents without converting to standard form?
Skill Practice
- Write the equation: center $(0, 6)$, radius $4\sqrt{2}$.
- Find center and radius: $(x + 9)^2 + (y - 3)^2 = 121$.
- Convert to standard form: $x^2 + y^2 - 14x + 2y + 14 = 0$.
- Convert to standard form: $3x^2 + 3y^2 + 18x - 12y + 9 = 0$.
- Find focus and directrix: $x^2 = -24y$.
- Find focus and directrix: $y^2 = 8x$.
- Write equation: vertex $(0,0)$, directrix $y = 3$.
- Write equation: vertex $(4, -2)$, focus $(4, 2)$.
- Find all features of $\dfrac{x^2}{100} + \dfrac{y^2}{64} = 1$.
- Find all features of $\dfrac{(x+5)^2}{49} + \dfrac{(y-1)^2}{24} = 1$.
- Write equation: center $(0,0)$, foci $(\pm 4, 0)$, $a = 7$.
- Write equation: center $(0,0)$, vertical major axis, $a = 9$, $b = 5$.
- Find all features of $\dfrac{x^2}{36} - \dfrac{y^2}{64} = 1$.
- Find all features of $\dfrac{(y+2)^2}{4} - \dfrac{(x-1)^2}{12} = 1$.
- Write equation: center $(0,0)$, vertices $(0, \pm 8)$, $b = 6$.
- Write equation: center $(0,0)$, vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$.
- Identify: $6x^2 + 6y^2 - 12x + 24y - 6 = 0$.
- Identify: $x^2 - 9y^2 + 2x + 18y - 9 = 0$.
- Identify and convert to standard form: $25x^2 + 4y^2 + 150x - 16y + 141 = 0$.
- Identify: $y^2 - 4y + 8x + 28 = 0$.
Word Problems
For each problem, define variables, set up the equation, and answer the specific question.
- A searchlight reflector is parabolic with the bulb 3 inches from the vertex. The reflector opening is 18 inches in diameter. How deep is the reflector?
- An elliptical city park will be 240 yards long and 160 yards wide. Two fountains will be placed at the foci. How far apart will the fountains be?
- The Moon's closest approach to Earth (perigee) is approximately 362,600 km and its farthest (apogee) is approximately 405,400 km. Find $a$, $c$, $b$, and the eccentricity of the Moon's orbit.
- A hyperbolic mirror is used in a reflecting telescope. Its equation in a coordinate model is $\dfrac{x^2}{16} - \dfrac{y^2}{20} = 1$. Find the center, vertices, foci, and asymptotes.
- A stone arch bridge spans a river with a semi-elliptical arch 60 meters wide and 20 meters tall at center. Write the equation of the ellipse and find the arch height 15 meters from center.
- A radio navigation system places two transmitters 2,000 meters apart. A ship's receiver determines it is always 1,200 meters closer to one transmitter than the other. Describe the shape of all possible ship positions and write an equation using appropriate coordinates.
Error Analysis
Each problem shows a student's work. Identify the error and write the correct solution.
Problem 1
A student rewrites $x^2 + y^2 + 4x - 6y = 12$ as $(x+2)^2 + (y-3)^2 = 12$.
What is wrong? What should the right side be?
Problem 2
A student identifies the vertices of $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$ as $(\pm 5, 0)$.
Explain the error. What are the correct vertices?
Problem 3
For a hyperbola with $a = 5$, $b = 12$, a student computes $c^2 = 25 - 144 = -119$ and stops.
Identify the error. Compute the correct value of $c$.
Problem 4
A student says $(x-3)^2 = 8(y+1)$ has its focus at $(3, 1)$.
What value of $p$ does this equation have? Where is the correct focus?
Problem 5
A student writes the asymptotes of $\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1$ as $y = \pm\dfrac{3}{4}x$.
Identify the error. What are the correct asymptote equations?
Chapter 12 Checkpoint
Chapter 12 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 12
Part A — Short Answer and Vocabulary
- State the standard form equation of a circle with center $(h, k)$ and radius $r$.
- What distance condition defines an ellipse?
- For a hyperbola with $a = 5$ and $b = 12$, find $c$.
- What does eccentricity measure, and what is the eccentricity of a circle?
Part B — Identification (no converting required)
- $4x^2 + 4y^2 - 8x + 16y + 4 = 0$
- $x^2 - 4y^2 + 6x + 8y + 1 = 0$
- $3x^2 + 5y^2 - 12x + 10y - 15 = 0$
- $x^2 + 6x + 4y - 3 = 0$
Part C — Standard Form and Features
- Rewrite $x^2 + y^2 - 6x + 8y - 11 = 0$ in standard form. State the center and radius.
- Find the center, vertices, co-vertices, and foci of $\dfrac{(x-2)^2}{25} + \dfrac{(y+3)^2}{16} = 1$.
- Find the center, vertices, foci, and asymptotes of $\dfrac{(y-1)^2}{9} - \dfrac{(x+4)^2}{16} = 1$.
- Write the equation of the parabola with vertex $(-1, 2)$ and directrix $y = 0$.
Part D — Applications
- A satellite dish is 3 meters wide and 0.75 meters deep, vertex at the origin. Write the equation of the parabolic cross-section and find the location of the focus (receiver).
- An elliptical conference table is 12 feet long and 6 feet wide. An audio engineer wants to install microphones at the foci so that speakers at one focus can be clearly heard at the other. How far from the center should the microphones be placed?
- A hyperbola has equation $\dfrac{x^2}{4} - \dfrac{y^2}{5} = 1$. Find all key features. Write the equations of the asymptotes.
Part E — Error Analysis
- A student rewrites $x^2 + y^2 + 4x - 6y = 12$ as $(x+2)^2 + (y-3)^2 = 12$. Identify the error and correct the equation.
- A student identifies the vertices of $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$ as $(\pm 5, 0)$. Identify the error. What are the correct vertices?
Answers may be placed in the answer key.
Bridge Note — Looking Ahead to Chapter 13
Chapter 13: Trigonometric Foundations
You are now prepared to move into Trigonometric Foundations. Trigonometry and conic sections are deeply connected in four important ways.
The unit circle. In Section 12.3, you wrote the standard equation of any circle as $(x-h)^2 + (y-k)^2 = r^2$. The unit circle is the special case $x^2 + y^2 = 1$ — a circle centered at the origin with radius 1. Chapter 13 introduces this circle as the foundation for defining sine and cosine for all angles.
Parametric equations of conics. The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ becomes $x = a\cos\theta$, $y = b\sin\theta$ in parametric form. For a circle, this simplifies to $x = r\cos\theta$, $y = r\sin\theta$. Trigonometric functions allow you to trace each conic curve as a point moves through angle values.
Polar equations of conics. All four conic sections unify into a single polar equation involving eccentricity — an elegant formula you cannot fully appreciate without both this chapter's coordinate geometry and Chapter 13's trigonometry.
Kepler's laws and calculus. In the MGU Calculus course, you will prove — using calculus, vectors, and trigonometry — why gravitational attraction causes planets to follow elliptical orbits. The ellipse you studied algebraically here reappears as the mathematically inevitable consequence of an inverse-square force law.
MGU Library Connection
- Conic Section Formula Reference — Appendix K of this textbook
- Algebra II Chapter 3: Quadratic Functions in Depth (parabola foundations)
- Algebra II Chapter 13: Trigonometric Foundations (unit circle, parametric equations)
- MGU Precalculus: Parametric Equations and Polar Conics
- MGU Physics: Classical Mechanics (Kepler's laws and orbital ellipses)
- MGU Calculus I: Volumes of Revolution and Arc Length of Conics
- MGU Dictionary entries: conic section, eccentricity, focus, asymptote, directrix