Conic Sections

Study Sheet

Conic Sections

Circles, parabolas, ellipses & hyperbolas --- with worked examples and labeled graphs

The Four Conics & Review Formulas

Concept
Big idea

A conic section is a curve you get by slicing a double cone with a flat plane. Tilting the plane a little more each time produces, in order, a circle, an ellipse, a parabola, and a hyperbola. Every one of them can be written as a single second-degree equation

Ax2+Cy2+Dx+Ey+F=0,Ax^2+Cy^2+Dx+Ey+F=0,

and the whole topic is really one question asked four ways: what shape does this equation draw, and where are its key points?

Concept
Two tools you already own (review)

For points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2):

Distance: d=(x2x1)2+(y2y1)2,Midpoint: M=(x1+x22, y1+y22).\textbf{Distance: } d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\qquad \textbf{Midpoint: } M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right).

The distance formula is the circle equation in disguise (a circle is all points a fixed distance from the center), so keep it close.

Example
Example: distance and midpoint

For (3,2)(-3,-2) and (5,4)(5,4):

d=(5(3))2+(4(2))2=82+62=100=10,d=\sqrt{(5-(-3))^2+(4-(-2))^2}=\sqrt{8^2+6^2}=\sqrt{100}=10,
M=(3+52,2+42)=(1,1).M=\left(\tfrac{-3+5}{2},\tfrac{-2+4}{2}\right)=(1,1).

Tip

Tip: A circle is just a special ellipse where both radii are equal. If you understand the ellipse deeply, the circle is free.

Circles

Concept
Standard equation of a circle

A circle with center (h,k)(h,k) and radius rr is the set of all points a distance rr from the center:

(xh)2+(yk)2=r2.(x-h)^2+(y-k)^2=r^2.

Read the center straight off the equation as the opposites of the numbers inside the parentheses, and take r=r2r=\sqrt{\,r^2\,}.

Example
Example: center and radius

For (x3)2+(y+2)2=25(x-3)^2+(y+2)^2=25 the center is (3,2)(3,-2) (note y+2=y(2)y+2=y-(-2)) and the radius is r=25=5r=\sqrt{25}=5.

Example
Example: completing the square (general form \to standard form)

Convert x2+y26x+4y12=0x^2+y^2-6x+4y-12=0 into standard form.

Group the xx's and yy's and move the constant:

(x26x)+(y2+4y)=12.(x^2-6x)+(y^2+4y)=12.

Complete each square by adding (62)2=9\left(\tfrac{-6}{2}\right)^2=9 and (42)2=4\left(\tfrac{4}{2}\right)^2=4 to both sides:

(x26x+9)+(y2+4y+4)=12+9+4,(x^2-6x+9)+(y^2+4y+4)=12+9+4,
(x3)2+(y+2)2=25.(x-3)^2+(y+2)^2=25.

So the center is (3,2)(3,-2) and the radius is r=5r=5.

Tip

Tip: When you complete the square, whatever you add on the left you must add on the right. A common slip is adding 99 and 44 only on one side.

Parabolas as Conics

Concept
Vertex form and direction of opening

A parabola is every point equidistant from a fixed focus and a fixed line called the directrix. In vertex form:

y=a(xh)2+kopens up (a>0) or down (a<0),y=a(x-h)^2+k \quad\text{opens up } (a>0)\text{ or down } (a<0),
x=a(yk)2+hopens right (a>0) or left (a<0).x=a(y-k)^2+h \quad\text{opens right } (a>0)\text{ or left } (a<0).

Either way the vertex is (h,k)(h,k): it is the turning point, halfway between the focus and the directrix.

Concept
Focus and directrix (intro)

Write an upward parabola as y=14p(xh)2+ky=\dfrac{1}{4p}(x-h)^2+k. Then pp is the distance from the vertex to the focus (and to the directrix):

focus (h,k+p),directrix y=kp.\textbf{focus }(h,\,k+p),\qquad \textbf{directrix } y=k-p.

The focus sits inside the curve; the directrix is the same distance out the other side.

Example
Example: vertex, focus, directrix

For y=14x2y=\dfrac{1}{4}x^2 we have 14p=14\dfrac{1}{4p}=\dfrac14, so 4p=44p=4 and p=1p=1. The vertex is (0,0)(0,0), the parabola opens up, the focus is (0,1)(0,1), and the directrix is the line y=1y=-1.

Tip

Tip: An x2x^2 term (and no y2y^2) means the parabola opens up or down; a y2y^2 term (and no x2x^2) means it opens left or right. A parabola is the only conic with just one squared variable.

Ellipses

Concept
Standard equation of an ellipse (center at origin)
x2a2+y2b2=1,a>b>0.\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\qquad a>b>0.

The larger denominator sits under the axis holding the major axis (the long one). Let aa be the larger of the two, bb the smaller. Then:

  • [leftmargin=*]
  • Vertices: endpoints of the major axis, distance aa from center.
  • Co-vertices: endpoints of the minor axis, distance bb from center.
  • Foci: distance cc from center along the major axis, where c2=a2b2c^2=a^2-b^2.

A shifted ellipse uses (xh)2a2+(yk)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1 with center (h,k)(h,k).

Example
Example: full feature analysis

Analyze x225+y29=1\dfrac{x^2}{25}+\dfrac{y^2}{9}=1.

The larger denominator (2525) is under x2x^2, so the major axis is horizontal with a=25=5a=\sqrt{25}=5 and b=9=3b=\sqrt{9}=3. Then c2=a2b2=259=16c^2=a^2-b^2=25-9=16, so c=4c=4.

  • [leftmargin=*]
  • Center (0,0)(0,0);   Vertices (±5,0)(\pm5,0);   Co-vertices (0,±3)(0,\pm3);   Foci (±4,0)(\pm4,0).
Tip

Tip: For an ellipse a>ba>b and the foci hug the long axis, so c<ac<a always. If you ever get c2<0c^2<0, you mislabeled aa and bb.

Hyperbolas

Concept
Standard equation of a hyperbola (center at origin)
x2a2y2b2=1  (opens left–right),y2a2x2b2=1  (opens up–down).\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \ \ (\text{opens left--right}),\qquad \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 \ \ (\text{opens up--down}).

The positive term tells you which way it opens; aa is always the number under the positive term.

  • [leftmargin=*]
  • Vertices: distance aa from center, on the opening axis.
  • Foci: distance cc from center, where c2=a2+b2c^2=a^2+b^2 (note the ++!).
  • Asymptotes: for x2a2y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 they are y=±baxy=\pm\dfrac{b}{a}x; for the up--down form they are y=±abxy=\pm\dfrac{a}{b}x.
Example
Example: full feature analysis

Analyze x29y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1.

The x2x^2 term is positive, so it opens left--right with a=9=3a=\sqrt{9}=3 and b=16=4b=\sqrt{16}=4. Then c2=a2+b2=9+16=25c^2=a^2+b^2=9+16=25, so c=5c=5.

  • [leftmargin=*]
  • Center (0,0)(0,0);   Vertices (±3,0)(\pm3,0);   Foci (±5,0)(\pm5,0);   Asymptotes y=±43xy=\pm\dfrac{4}{3}x.
Tip

Tip: Ellipse uses c2=a2b2c^2=a^2-b^2; hyperbola uses c2=a2+b2c^2=a^2+b^2. For a hyperbola the foci are farther out than the vertices, so c>ac>a.

Identifying a Conic from General Form

Concept
The classification test

Given Ax2+Cy2+Dx+Ey+F=0Ax^2+Cy^2+Dx+Ey+F=0 (no xyxy term), look only at AA and CC:

  • [leftmargin=*]
  • Circle: A=CA=C (same coefficient on x2x^2 and y2y^2).
  • Parabola: exactly one of A,CA,C is 00 (only one squared term).
  • Ellipse: ACA\ne C but AA and CC have the same sign.
  • Hyperbola: AA and CC have opposite signs.
Example
Example: classify each
  • [leftmargin=*]
  • x2+y24x=1x^2+y^2-4x=1: A=C=1A=C=1 \Rightarrow circle.
  • y=3x2xy=3x^2-x (only x2x^2): parabola.
  • 4x2+9y2=364x^2+9y^2=36: A=4,C=9A=4,\,C=9 same sign, unequal \Rightarrow ellipse.
  • 9x24y2=369x^2-4y^2=36: A=9,C=4A=9,\,C=-4 opposite signs \Rightarrow hyperbola.
Tip

Tip (how to tell them apart at a glance): One squared term \Rightarrow parabola. Two squared terms: same number \Rightarrow circle, added \Rightarrow ellipse, subtracted \Rightarrow hyperbola.

Applications (light)

Concept
Where conics show up
  • [leftmargin=*]
  • Ellipses --- orbits: planets and satellites travel in ellipses with the larger body at one focus (Kepler's First Law).
  • Ellipses --- whispering galleries: a sound starting at one focus reflects off the elliptical ceiling and gathers at the other focus, so a whisper carries across the room.
  • Parabolas --- dishes and headlights: a satellite dish or reflector collects incoming rays at its focus (and a bulb at the focus sends out a straight beam).
  • Hyperbolas --- navigation and comets: some comets swing past the sun on an open hyperbolic path and never return.
Tip

Tip: The focus is the star of every application. Ellipses and hyperbolas have two foci; a parabola has one. That reflecting property is exactly why the focus matters.

Going Deeper: Advanced Conics

Concept
Building a conic from conditions

So far you have read a conic off its equation. The reverse skill --- constructing the equation from geometric conditions --- is what real problems ask for. Each unknown coefficient needs one condition:

  • [leftmargin=*]
  • Through given points: plug each point into the general form and solve the resulting linear system for the coefficients.
  • Given foci and eccentricity: the foci fix the center and cc; then e=cae=\dfrac{c}{a} gives aa, and bb comes from b2=a2c2b^2=a^2-c^2 (ellipse) or b2=c2a2b^2=c^2-a^2 (hyperbola).
  • Given a tangent line: impose the tangency condition (next box) as one more equation.

Eccentricity ee measures how “stretched” a conic is: e=0e=0 circle, 0<e<10<e<1 ellipse, e=1e=1 parabola, e>1e>1 hyperbola.

Example
Example: ellipse from foci and eccentricity

Find the ellipse centered at the origin with foci (±3,0)(\pm3,0) and eccentricity e=35e=\dfrac35.

The foci are on the xx-axis at distance c=3c=3, so the major axis is horizontal. From e=cae=\dfrac{c}{a},

a=ce=33/5=5,a2=25.a=\frac{c}{e}=\frac{3}{3/5}=5,\qquad a^2=25.

Recompute bb using c2=a2b2c^2=a^2-b^2:

b2=a2c2=259=16.b^2=a^2-c^2=25-9=16.

So the ellipse is

x225+y216=1,\frac{x^2}{25}+\frac{y^2}{16}=1,

with vertices (±5,0)(\pm5,0), co-vertices (0,±4)(0,\pm4), and foci (±3,0)(\pm3,0) as required.

Concept
The line--conic tangency condition (discriminant =0=0)

To find where a line meets a conic, substitute the line into the conic. You get a quadratic in one variable, so the discriminant Δ=B24AC\Delta=B^2-4AC of that quadratic tells you everything:

Δ>0  2 points (secant),Δ=0  1 point (tangent),Δ<0  no intersection.\Delta>0 \ \Rightarrow\ 2 \text{ points (secant)},\quad \Delta=0 \ \Rightarrow\ 1 \text{ point (tangent)},\quad \Delta<0 \ \Rightarrow\ \text{no intersection}.

So “the line is tangent to the conic” translates into the single algebraic equation Δ=0\Delta=0. For the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 this collapses to the tidy rule: the line y=mx+ky=mx+k is tangent exactly when

k2=a2m2+b2.k^2=a^2m^2+b^2.
Example
Example: tangent lines by the discriminant

For which values of kk is the line y=x+ky=x+k tangent to x24+y2=1\dfrac{x^2}{4}+y^2=1?

Substitute y=x+ky=x+k and clear the fraction (multiply by 44):

x24+(x+k)2=1  x2+4(x+k)2=4.\frac{x^2}{4}+(x+k)^2=1 \ \Longrightarrow\ x^2+4(x+k)^2=4.

Expand into a quadratic in xx:

5x2+8kx+(4k24)=0.5x^2+8kx+(4k^2-4)=0.

Set its discriminant to zero:

Δ=(8k)24(5)(4k24)=64k280k2+80=16k2+80=0,\Delta=(8k)^2-4(5)(4k^2-4)=64k^2-80k^2+80=-16k^2+80=0,

so k2=5k^2=5 and k=±5k=\pm\sqrt5. Checking against the shortcut with a2=4a^2=4, b2=1b^2=1, m=1m=1: k2=a2m2+b2=4+1=5k^2=a^2m^2+b^2=4+1=5. ✓

Concept
Latus rectum, focal chords, and the reflection property

A focal chord is any chord through a focus. The special focal chord perpendicular to the major (or focal) axis is the latus rectum, and its length measures how “wide” the curve is at the focus:

ellipse / hyperbola:  2b2a,parabola y=14px2: 4p.\text{ellipse / hyperbola: } \ \frac{2b^2}{a},\qquad \text{parabola } y=\tfrac{1}{4p}x^2: \ 4p.

The reflection property explains every application from Section 7: a ray aimed at one focus of an ellipse bounces to the other focus; a ray parallel to a parabola's axis reflects through its single focus; and a ray toward one focus of a hyperbola reflects away along the line to the other focus.

Example
Example: latus rectum of an ellipse

For x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1 we have a=5a=5, b=4b=4, so each latus rectum has length

2b2a=2(16)5=325=6.4.\frac{2b^2}{a}=\frac{2(16)}{5}=\frac{32}{5}=6.4 .

Since the foci are (±3,0)(\pm3,0), the latus-rectum endpoints above and below the right focus are (3,±165)=(3,±3.2)\left(3,\pm\dfrac{16}{5}\right)=(3,\pm3.2).

Concept
Classifying the full second-degree equation via B24ACB^2-4AC

When an xyxy term is present the conic is rotated, and reading AA vs. CC is no longer enough. The general equation

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2+Bxy+Cy^2+Dx+Ey+F=0

is classified by the invariant B24ACB^2-4AC:

B24AC<0  ellipse (circle if A=C, B=0),B^2-4AC<0 \ \Rightarrow\ \textbf{ellipse}\ (\text{circle if }A=C,\ B=0),
B24AC=0  parabola,B24AC>0  hyperbola.B^2-4AC=0 \ \Rightarrow\ \textbf{parabola},\qquad B^2-4AC>0 \ \Rightarrow\ \textbf{hyperbola}.

This quantity is unchanged by any rotation of the axes, which is why it works even when B0B\ne0. (When B=0B=0 it agrees with the sign test of Section 6.)

Example
Example: classify a rotated conic

Classify 2x2+3xy+2y25=02x^2+3xy+2y^2-5=0. Here A=2A=2, B=3B=3, C=2C=2, so

B24AC=324(2)(2)=916=7<0,B^2-4AC=3^2-4(2)(2)=9-16=-7<0,

an ellipse (tilted, since B0B\ne0). By contrast x24xy+4y2+x=0x^2-4xy+4y^2+x=0 gives B24AC=164(1)(4)=0B^2-4AC=16-4(1)(4)=0, a parabola.

Concept
Degenerate conics

A “conic” equation does not always draw a smooth curve. When the slicing plane passes through the apex of the cone you get a degenerate conic:

  • [leftmargin=*]
  • A single point --- e.g. x2+y2=0x^2+y^2=0 is only (0,0)(0,0) (a “circle of radius 00”).
  • One line / two parallel lines --- e.g. x2=4x^2=4 gives x=±2x=\pm2 (a degenerate parabola).
  • Two intersecting lines --- e.g. x2y2=0x^2-y^2=0 gives y=±xy=\pm x (a degenerate hyperbola).
  • Empty set (no points) --- e.g. x2+y2=1x^2+y^2=-1 has no real solutions.

Watch for these after completing the square: a right-hand side of 00 or a negative number is the tell.

Concept
Shortest distance from a point to a conic

The closest point QQ on a conic to an external point PP has a clean geometric signature: the segment PQPQ is perpendicular to the tangent line at QQ (it lies along the normal). For a circle this makes the answer immediate --- the nearest point lies on the line through PP and the center:

distance from P to circle=PCr,\text{distance from }P\text{ to circle}=\bigl|\,\overline{PC}-r\,\bigr|,

where CC is the center and rr the radius. For other conics you either minimize the squared distance d2d^2 with calculus or impose the normal condition and solve.

Example
Example: point to a circle

Find the shortest distance from P=(8,0)P=(8,0) to the circle (x2)2+y2=9(x-2)^2+y^2=9.

The center is C=(2,0)C=(2,0) and r=3r=3. The distance from PP to the center is

PC=(82)2+(00)2=36=6.\overline{PC}=\sqrt{(8-2)^2+(0-0)^2}=\sqrt{36}=6.

Since PP is outside the circle (6>36>3), the shortest distance is

PCr=63=3,\bigl|\overline{PC}-r\bigr|=|6-3|=3,

reached at the point (5,0)(5,0) where the segment PCPC crosses the circle.

Tip

Tip: Discriminants run this whole section. B24ACB^2-4AC (from the equation) tells you which conic; the discriminant of the substituted quadratic (from a line-plus-conic) tells you how a line meets it. Same tool, two jobs.

Formulas, Proofs & Tips

Tip
Equation of a circle
(xh)2+(yk)2=r2(x-h)^2+(y-k)^2=r^2

What it means. All points at distance rr from the centre (h,k)(h,k).

Example. Center (2,1)(2,-1), radius 33: (x2)2+(y+1)2=9(x-2)^2+(y+1)^2=9.

Why it works. A circle is by definition the set of points a fixed distance from the centre. Writing that distance with the distance formula gives (xh)2+(yk)2=r\sqrt{(x-h)^2+(y-k)^2}=r; squaring both sides removes the root.

Tip. Given x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0, complete the square in xx and in yy to recover the centre and radius.