Conic Sections and Parametric Equations

Study Sheet

Conic Sections and Parametric Equations

Parabolas, ellipses, hyperbolas, classifying by completing the square, parametric curves, and polar conics

Parabolas: Focus, Directrix, and Latus Rectum

Concept
Standard forms of a parabola (vertex (h,k)(h,k))

A parabola is the set of points equidistant from a fixed focus and a fixed line, the directrix. The number pp is the signed distance from vertex to focus.

(xh)2=4p(yk)opens up/down,(yk)2=4p(xh)opens right/left.(x-h)^2=4p(y-k)\quad\text{opens up/down},\qquad (y-k)^2=4p(x-h)\quad\text{opens right/left}.
  • [leftmargin=5mm]
  • Vertical axis: focus (h,k+p)(h,\,k+p), directrix y=kpy=k-p. Opens up if p>0p>0, down if p<0p<0.
  • Horizontal axis: focus (h+p,k)(h+p,\,k), directrix x=hpx=h-p. Opens right if p>0p>0, left if p<0p<0.
  • Latus rectum (focal chord through the focus, perpendicular to the axis) has length 4p|4p|.
  • Eccentricity of every parabola is e=1e=1.
Example
Completing the square: x26x8y+1=0x^2-6x-8y+1=0

Group the squared variable and complete the square:

x26x=8y1x26x+9=8y1+9(x3)2=8(y+1).\begin{aligned} x^2-6x &= 8y-1\\ x^2-6x+9 &= 8y-1+9\\ (x-3)^2 &= 8(y+1). \end{aligned}

So 4p=8p=24p=8\Rightarrow p=2; vertex (3,1)(3,-1); opens up. Focus (3,1+2)=(3,1)(3,\,-1+2)=(3,1); directrix y=12=3y=-1-2=-3; latus rectum =4p=8=|4p|=8.

Example
From standard form: y2=12xy^2=12x

Horizontal axis, vertex (0,0)(0,0). 4p=12p=34p=12\Rightarrow p=3, opens right. Focus (3,0)(3,0); directrix x=3x=-3; latus rectum =12=12.

Tip

Tip: The squared variable tells you the axis. If xx is squared, the parabola opens up/down; if yy is squared, it opens left/right. The focus always sits inside the curve, the directrix outside.

Ellipses: Center, Axes, Foci, Eccentricity

Concept
Standard forms of an ellipse (center (h,k)(h,k), a>b>0a>b>0)
(xh)2a2+(yk)2b2=1(major axis horizontal),(xh)2b2+(yk)2a2=1(major axis vertical).\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\quad\text{(major axis horizontal)},\qquad \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\quad\text{(major axis vertical)}.
  • [leftmargin=5mm]
  • aa = semi-major axis (largest denominator, a2a^2); bb = semi-minor axis.
  • Vertices are aa from center along the major axis; co-vertices are bb from center along the minor axis.
  • Foci lie on the major axis, cc from center, where c2=a2b2c^2=a^2-b^2.
  • Eccentricity e=cae=\dfrac{c}{a}, with 0<e<10<e<1 (nearer 00 is more circular).
Example
Completing the square: 4x2+9y216x+18y11=04x^2+9y^2-16x+18y-11=0

Group by variable, factor out leading coefficients, then complete each square:

4(x24x)+9(y2+2y)=114(x24x+4)+9(y2+2y+1)=11+16+94(x2)2+9(y+1)2=36(x2)29+(y+1)24=1.\begin{aligned} 4(x^2-4x)+9(y^2+2y) &= 11\\ 4(x^2-4x+4)+9(y^2+2y+1) &= 11+16+9\\ 4(x-2)^2+9(y+1)^2 &= 36\\ \frac{(x-2)^2}{9}+\frac{(y+1)^2}{4} &= 1. \end{aligned}

Center (2,1)(2,-1); a2=9, b2=4a^2=9,\ b^2=4 so a=3, b=2a=3,\ b=2; major axis horizontal. c2=94=5c=5c^2=9-4=5\Rightarrow c=\sqrt5. Vertices (5,1),(1,1)(5,-1),(-1,-1); co-vertices (2,1),(2,3)(2,1),(2,-3); foci (2±5,1)(2\pm\sqrt5,\,-1); e=530.75e=\dfrac{\sqrt5}{3}\approx0.75.

Tip

Tip: The larger denominator is always a2a^2, and it names the direction of the major axis: under xx means horizontal, under yy means vertical. Add the amounts you completed to both sides, remembering to multiply by the factored-out coefficient.

Hyperbolas: Vertices, Foci, Asymptotes, Eccentricity

Concept
Standard forms of a hyperbola (center (h,k)(h,k))
(xh)2a2(yk)2b2=1(opens left/right),(yk)2a2(xh)2b2=1(opens up/down).\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\quad\text{(opens left/right)},\qquad \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\quad\text{(opens up/down)}.
  • [leftmargin=5mm]
  • The positive term names the transverse axis; a2a^2 is its denominator (it need not be the larger one).
  • Vertices are aa from center along the transverse axis; foci are cc from center with c2=a2+b2c^2=a^2+b^2.
  • Asymptotes: yk=±ba(xh)y-k=\pm\dfrac{b}{a}(x-h) for the left/right form; yk=±ab(xh)y-k=\pm\dfrac{a}{b}(x-h) for the up/down form.
  • Eccentricity e=cae=\dfrac{c}{a}, with e>1e>1.
Example
Completing the square: 4x29y2+8x+36y68=04x^2-9y^2+8x+36y-68=0

Watch the sign on the yy group when factoring out 9-9:

4(x2+2x)9(y24y)=684(x2+2x+1)9(y24y+4)=68+4364(x+1)29(y2)2=36(x+1)29(y2)24=1.\begin{aligned} 4(x^2+2x)-9(y^2-4y) &= 68\\ 4(x^2+2x+1)-9(y^2-4y+4) &= 68+4-36\\ 4(x+1)^2-9(y-2)^2 &= 36\\ \frac{(x+1)^2}{9}-\frac{(y-2)^2}{4} &= 1. \end{aligned}

Center (1,2)(-1,2); a2=9, b2=4a^2=9,\ b^2=4 so a=3, b=2a=3,\ b=2; opens left/right. c2=9+4=13c=13c^2=9+4=13\Rightarrow c=\sqrt{13}. Vertices (2,2),(4,2)(2,2),(-4,2); foci (1±13,2)(-1\pm\sqrt{13},\,2); asymptotes y2=±23(x+1)y-2=\pm\dfrac{2}{3}(x+1); e=1331.20e=\dfrac{\sqrt{13}}{3}\approx1.20.

Tip

Tip: For a hyperbola cc is the largest of the three (c>ac>a and c>bc>b) because c2=a2+b2c^2=a^2+b^2. Do not assume a>ba>b: for a hyperbola aa is simply whichever denominator sits under the positive term.

Classifying Conics from the General Form

Concept
The general second-degree equation

With no xyxy term, Ax2+Cy2+Dx+Ey+F=0Ax^2+Cy^2+Dx+Ey+F=0 classifies by comparing AA and CC:

  • [leftmargin=5mm]
  • Circle: A=CA=C (and same sign), e.g. A=C0A=C\ne0.
  • Parabola: exactly one of A,CA,C is 00 (only one variable is squared).
  • Ellipse: AA and CC have the same sign but ACA\ne C.
  • Hyperbola: AA and CC have opposite signs (AC<0AC<0).
Example
Classify each equation
  • [leftmargin=5mm]
  • 3x2+3y212x+6=03x^2+3y^2-12x+6=0: A=C=3A=C=3\Rightarrow circle.
  • x28y+2x=0x^2-8y+2x=0: only xx is squared \Rightarrow parabola.
  • 2x2+5y24=02x^2+5y^2-4=0: same sign, 252\ne5\Rightarrow ellipse.
  • 9x24y2+18x=09x^2-4y^2+18x=0: opposite signs \Rightarrow hyperbola.
Tip

Tip: Classify before completing the square. The sign comparison of AA and CC instantly names the conic; completing the square then locates its center and features.

Parametric Equations

Concept
Curves defined by a parameter

A parametric curve gives xx and yy each as a function of a third variable tt:

x=f(t),y=g(t).x=f(t),\qquad y=g(t).

As tt increases the point (x,y)(x,y) traces the curve in a definite direction (its orientation).

  • [leftmargin=5mm]
  • Graph by a table: choose values of tt, compute (x,y)(x,y), and connect in order of increasing tt.
  • Eliminate the parameter: solve one equation for tt (or use a trig identity) and substitute to get a relation in xx and yy.
Example
Eliminate the parameter (polynomial): x=t2, y=tx=t^2,\ y=t

From the second equation t=yt=y. Substitute: x=y2x=y^2. The curve is a sideways parabola opening right, traced upward as tt increases. Building a table:

Example
Eliminate the parameter (trig): x=3cost, y=2sintx=3\cos t,\ y=2\sin t

Solve for the trig functions: cost=x3\cos t=\dfrac{x}{3} and sint=y2\sin t=\dfrac{y}{2}. Apply cos2t+sin2t=1\cos^2 t+\sin^2 t=1:

(x3)2+(y2)2=1x29+y24=1.\left(\frac{x}{3}\right)^2+\left(\frac{y}{2}\right)^2=1\quad\Longrightarrow\quad \frac{x^2}{9}+\frac{y^2}{4}=1.

This is an ellipse with a=3, b=2a=3,\ b=2, traced counterclockwise.

Example
Writing a parametrization

To parametrize the line segment from (1,2)(1,2) to (5,4)(5,-4) for 0t10\le t\le1, use

x=1+(51)t=1+4t,y=2+(42)t=26t.x=1+(5-1)t=1+4t,\qquad y=2+(-4-2)t=2-6t.

At t=0t=0 you are at (1,2)(1,2); at t=1t=1 at (5,4)(5,-4).

Tip

Tip: For trig parametrizations reach for the identity cos2t+sin2t=1\cos^2 t+\sin^2 t=1 (or sec2ttan2t=1\sec^2 t-\tan^2 t=1 for hyperbolas) rather than solving for tt. Always state the direction and any restriction on the resulting graph.

Polar Equations of Conics (Introduction)

Concept
Eccentricity form

With a focus at the pole, a conic has polar equation

r=ed1±ecosθorr=ed1±esinθ,r=\frac{ed}{1\pm e\cos\theta}\qquad\text{or}\qquad r=\frac{ed}{1\pm e\sin\theta},

where ee is the eccentricity and d>0d>0 is the distance from focus to directrix.

  • [leftmargin=5mm]
  • e=1e=1\Rightarrow parabola,   0<e<10<e<1\Rightarrow ellipse,   e>1e>1\Rightarrow hyperbola.
  • cos\cos means the directrix is vertical; sin\sin means it is horizontal.
Example
Identify r=62+cosθr=\dfrac{6}{2+\cos\theta}

Put the constant term to 11 by dividing numerator and denominator by 22:

r=62+cosθ=31+12cosθ.r=\frac{6}{2+\cos\theta}=\frac{3}{1+\tfrac12\cos\theta}.

So e=12<1e=\tfrac12<1: an ellipse. From ed=3ed=3 and e=12e=\tfrac12 we get d=6d=6; the directrix is vertical.

Tip

Tip: Before reading off ee, force the denominator into the form 1±e(cos or sin)θ1\pm e(\cos\text{ or }\sin)\theta by dividing through by the leading constant. Only then is the coefficient of cosθ\cos\theta or sinθ\sin\theta equal to ee.

Going Deeper: Advanced Conics & Parametrics

Concept
The line--conic tangency condition

Substitute a line y=mx+cy=mx+c into a conic and collect a quadratic in xx. The line is tangent exactly when that quadratic has a double root, i.e. its discriminant is 00 (secant if the discriminant is positive, misses if negative). Carrying this out once gives memorable tangency conditions for lines of slope mm:

  • [leftmargin=5mm]
  • Ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1:  tangent iff c2=a2m2+b2c^2=a^2m^2+b^2.
  • Hyperbola x2a2y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1:  tangent iff c2=a2m2b2c^2=a^2m^2-b^2.
  • Parabola y2=4pxy^2=4px:  tangent iff c=pmc=\dfrac{p}{m} (line y=mx+pmy=mx+\tfrac{p}{m}).

These same relations let you find a conic from tangency, point, or focus data: each condition gives one equation in the unknowns a2,b2,p,a^2,b^2,p,\dots, and you solve the resulting system.

Example
Worked: tangent lines of slope 22 to x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=1

Here a2=9, b2=4, m=2a^2=9,\ b^2=4,\ m=2. The tangency condition c2=a2m2+b2c^2=a^2m^2+b^2 gives

c2=9(2)2+4=40c=±210.c^2=9(2)^2+4=40\quad\Longrightarrow\quad c=\pm2\sqrt{10}.

So the two tangent lines of slope 22 are y=2x+210y=2x+2\sqrt{10} and y=2x210y=2x-2\sqrt{10}. (Check: substituting either into the ellipse yields a perfect-square quadratic in xx.)

Example
Worked: build a conic from a focus and a tangency condition

Find the ellipse centered at the origin with horizontal major axis, a focus at (3,0)(3,0), that is tangent to the line y=x+5y=x+5.

focus condition:c=3a2b2=c2=9,tangency (m=1, c=5):c2=a2m2+b225=a2+b2.\begin{aligned} \text{focus condition:}\quad c=3 &\Rightarrow a^2-b^2=c^2=9,\\ \text{tangency }(m=1,\ c=5):\quad c^2=a^2m^2+b^2 &\Rightarrow 25=a^2+b^2. \end{aligned}

Adding and subtracting: a2=17, b2=8a^2=17,\ b^2=8. The ellipse is x217+y28=1\dfrac{x^2}{17}+\dfrac{y^2}{8}=1. Since a2b2=9a^2-b^2=9, the foci are indeed (±3,0)(\pm3,0), confirming the fit.

Tip

Reflection property (why conics focus light and sound):

  • [leftmargin=5mm]
  • Parabola: rays traveling parallel to the axis reflect through the focus (and vice versa) --- the principle behind satellite dishes and headlights.
  • Ellipse: a ray leaving one focus reflects to the other focus --- “whispering galleries.”
  • Hyperbola: a ray aimed at one focus reflects away from the other focus.
Concept
Focal chords and the latus rectum

A focal chord is any chord through a focus. The latus rectum is the focal chord perpendicular to the major/transverse axis; its half-length \ell (the semi-latus rectum) controls the polar form below.

parabola y2=4px: L.R.=4p;ellipse & hyperbola: L.R.=2b2a,=b2a.\text{parabola }y^2=4px:\ \text{L.R.}=|4p|;\qquad \text{ellipse \& hyperbola:}\ \text{L.R.}=\frac{2b^2}{a},\quad \ell=\frac{b^2}{a}.

For an ellipse =a(1e2)\ell=a(1-e^2) and for a hyperbola =a(e21)\ell=a(e^2-1), so knowing aa and ee recovers the full shape. Every focal chord of a conic has semi-latus rectum \ell as its harmonic-mean radius.

Concept
Classifying a general conic by B24ACB^2-4AC

The full second-degree equation

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

may include a cross term BxyBxy, which rotates the axes. The rotation-invariant discriminant B24ACB^2-4AC classifies the curve without ever finding the rotation angle:

  • [leftmargin=5mm]
  • B24AC<0B^2-4AC<0\Rightarrow ellipse (circle if also A=C, B=0A=C,\ B=0),
  • B24AC=0B^2-4AC=0\Rightarrow parabola,
  • B24AC>0B^2-4AC>0\Rightarrow hyperbola.

When B=0B=0 this reduces to the earlier sign test on AA and CC. Shortest distance to a conic: the closest point of a smooth conic to an external line lies where a tangent is parallel to that line, so slide the tangency condition to the tangent line nearest the given one, then measure the gap between the two parallel lines.

Example
Worked: classify rotated conics
  • [leftmargin=5mm]
  • xy=1xy=1: here A=C=0, B=1A=C=0,\ B=1, so B24AC=1>0B^2-4AC=1>0\Rightarrow hyperbola (rotated 4545^\circ).
  • 5x24xy+8y236=05x^2-4xy+8y^2-36=0: B24AC=(4)24(5)(8)=16160=144<0B^2-4AC=(-4)^2-4(5)(8)=16-160=-144<0\Rightarrow ellipse.
  • x2+23xy+3y28x=0x^2+2\sqrt3\,xy+3y^2-8x=0: B24AC=(23)24(1)(3)=1212=0B^2-4AC=(2\sqrt3)^2-4(1)(3)=12-12=0\Rightarrow parabola.
Example
Worked: shortest distance from a line to an ellipse

Find the shortest distance from the line x+y=6x+y=6 to the ellipse x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=1. The line has slope m=1m=-1. The parallel tangents satisfy c2=a2m2+b2=9+4=13c^2=a^2m^2+b^2=9+4=13, so c=±13c=\pm\sqrt{13}; the one nearer the line is y=x+13y=-x+\sqrt{13}, i.e. x+y=13x+y=\sqrt{13}. The gap between the parallel lines x+y6=0x+y-6=0 and x+y13=0x+y-\sqrt{13}=0 is

d=61312+12=61322.3941.4141.69.d=\frac{\bigl|6-\sqrt{13}\bigr|}{\sqrt{1^2+1^2}}=\frac{6-\sqrt{13}}{\sqrt2}\approx\frac{2.394}{1.414}\approx1.69.
Example
Worked: eliminate the parameter (rational parametrization)

Consider x=t+1t, y=t1tx=t+\dfrac{1}{t},\ y=t-\dfrac{1}{t} for t0t\ne0. Rather than solving for tt, combine the equations: x+y=2tx+y=2t and xy=2tx-y=\dfrac{2}{t}, so their product removes tt:

(x+y)(xy)=(2t) ⁣(2t)=4x2y2=4.(x+y)(x-y)=\left(2t\right)\!\left(\frac{2}{t}\right)=4\quad\Longrightarrow\quad x^2-y^2=4.

This is a hyperbola with a=b=2a=b=2. Because x=t+1tx=t+\tfrac1t satisfies x2|x|\ge2, only the outer branches are traced: t>0t>0 gives the right branch, t<0t<0 the left. The same trick with x=asect, y=btantx=a\sec t,\ y=b\tan t uses sec2ttan2t=1\sec^2t-\tan^2t=1 to yield x2a2y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.

Tip

Polar eccentricity form, revisited: in r=1±ecosθr=\dfrac{\ell}{1\pm e\cos\theta} the numerator is the semi-latus rectum =ed\ell=ed, so once you normalize the denominator to 1±e(cos or sin)θ1\pm e(\cos\text{ or }\sin)\theta you can read off both ee (the coefficient) and \ell (the numerator), then recover a=1e2a=\dfrac{\ell}{1-e^2} for an ellipse or a=e21a=\dfrac{\ell}{e^2-1} for a hyperbola. Pitfalls: always normalize the leading denominator constant to 11 before reading ee; when a cross term is present, classify with B24ACB^2-4AC (not the AA-vs-CC test); and for rational or trig parametrizations always state the branch or orientation restriction the algebra hides.

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.