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.

(x−h)2=4p(y−k)opens up/down,(y−k)2=4p(x−h)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}.
  • Vertical axis: focus (h, k+p)(h,\,k+p), directrix y=k−py=k-p. Opens up if p>0p>0, down if p<0p<0.
  • Horizontal axis: focus (h+p, k)(h+p,\,k), directrix x=h−px=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.

Put the vertex at the origin, the focus at (0,p)(0,p) and the directrix at y=−py=-p. A point (x,y)(x,y) is equally far from both when x2+(y−p)2=y+p\sqrt{x^2+(y-p)^2}=y+p; squaring and cancelling leaves x2=4pyx^2=4py. So pp is not a stretch factor chosen for convenience: it is the actual distance from the vertex to the focus.

Example
Completing the square: x2−6x−8y+1=0x^2-6x-8y+1=0
F(3, 1)V(3, −1)y = −3
Every point, like the one at x=7x=7, is as far from the focus as from the directrix.

Group the squared variable and complete the square:

x2−6x=8y−1x2−6x+9=8y−1+9(x−3)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=8⇒p=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=−1−2=−3y=-1-2=-3; latus rectum =∣4p∣=8=|4p|=8.

Side note
Where it goes wrong

Eliminating the parameter and forgetting the restrictions it carried. x=cos⁡tx = \cos t, y=sin⁡ty = \sin t traces the circle once per 2π2\pi; x=t2x = t^2, y=ty = t is only the right half of the sideways parabola… actually all of it, but x=tx = \sqrt t would be only part.

Example
From standard form: y2=12xy^2=12x
F(3, 0)x = −3latus rectum 12
Opens right; the chord through the focus is 4p=124p=12 long.

Horizontal axis, vertex (0,0)(0,0). 4p=12⇒p=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.

Try it
Quick check

For x2=8yx^2=8y, give the vertex, focus, directrix, direction of opening, and latus rectum length.

Work it on paper before revealing the answer.

Show the answer

Answer: Vertex (0,0)(0,0); 4p=8, p=24p=8,\ p=2; focus (0,2)(0,2); directrix y=−2y=-2; opens up; latus rectum =8=8.

Ellipses: Center, Axes, Foci, Eccentricity

Concept
Standard forms of an ellipse (center (h,k)(h,k), a>b>0a>b>0)
(x−h)2a2+(y−k)2b2=1(major axis horizontal),(x−h)2b2+(y−k)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)}.
  • 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=a2−b2c^2=a^2-b^2.
  • Eccentricity e=cae=\dfrac{c}{a}, with 0<e<10<e<1 (nearer 00 is more circular).

An ellipse is the set of points whose distances to two foci add to the same total, 2a2a. At the end of the minor axis the two distances are equal, each aa, and they form a right triangle with legs bb and cc; that triangle is the relation a2=b2+c2a^2=b^2+c^2. The closer the foci sit to the centre (small cc), the rounder the ellipse, which is what e=cae=\tfrac ca measures.

Example
Completing the square: 4x2+9y2−16x+18y−11=04x^2+9y^2-16x+18y-11=0
F₁F₂
Centre (2,−1)(2,-1), a=3a=3, b=2b=2, foci 2±52\pm\sqrt5: the two dashed distances always add to 66.

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

4(x2−4x)+9(y2+2y)=114(x2−4x+4)+9(y2+2y+1)=11+16+94(x−2)2+9(y+1)2=36(x−2)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=9−4=5⇒c=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=53≈0.75e=\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.

Try it
Quick check

For x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1, give the vertices, co-vertices, foci and eccentricity.

Show the answer

Answer: vertices (±4,0)(\pm4,0), co-vertices (0,±3)(0,\pm3), foci (±7,0)(\pm\sqrt7,0), e=74e=\dfrac{\sqrt7}{4}. a=4a=4, b=3b=3, and c2=a2−b2=7c^2=a^2-b^2=7.

Hyperbolas: Vertices, Foci, Asymptotes, Eccentricity

Concept
Standard forms of a hyperbola (center (h,k)(h,k))
(x−h)2a2−(y−k)2b2=1(opens left/right),(y−k)2a2−(x−h)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)}.
  • 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: y−k=±ba(x−h)y-k=\pm\dfrac{b}{a}(x-h) for the left/right form; y−k=±ab(x−h)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: 4x2−9y2+8x+36y−68=04x^2-9y^2+8x+36y-68=0
FF
Centre (−1,2)(-1,2), asymptotes of slope ±23\pm\tfrac23, foci −1±13-1\pm\sqrt{13}.

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

4(x2+2x)−9(y2−4y)=684(x2+2x+1)−9(y2−4y+4)=68+4−364(x+1)2−9(y−2)2=36(x+1)29−(y−2)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=13⇒c=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 y−2=±23(x+1)y-2=\pm\dfrac{2}{3}(x+1); e=133≈1.20e=\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.

Try it
Quick check

For x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1, give the vertices, foci, asymptotes and eccentricity.

Show the answer

Answer: vertices (±4,0)(\pm4,0), foci (±5,0)(\pm5,0), asymptotes y=±34xy=\pm\tfrac34x, e=54e=\tfrac54. For a hyperbola c2=a2+b2=16+9=25c^2=a^2+b^2=16+9=25.

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:

  • Circle: A=CA=C (and same sign), e.g. A=C≠0A=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 A≠CA\ne C.
  • Hyperbola: AA and CC have opposite signs (AC<0AC<0).
Example
Classify each equation
  • 3x2+3y2−12x+6=03x^2+3y^2-12x+6=0: A=C=3⇒A=C=3\Rightarrow circle.
  • x2−8y+2x=0x^2-8y+2x=0: only xx is squared ⇒\Rightarrow parabola.
  • 2x2+5y2−4=02x^2+5y^2-4=0: same sign, 2≠5⇒2\ne5\Rightarrow ellipse.
  • 9x2−4y2+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.

Try it
Quick check

Classify: (a) 4x2−y2+8x+2y=04x^2-y^2+8x+2y=0; (b) x2+y2−4x+6y=3x^2+y^2-4x+6y=3; (c) y2−4x+2y=0y^2-4x+2y=0.

Show the answer

Answer: (a) hyperbola (AA, CC opposite signs); (b) circle (A=CA=C); (c) parabola (only yy is squared).

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).

  • 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.

A parametrization is a record of motion: tt is the clock and (f(t),g(t))(f(t),g(t)) is where the point is at that moment. Eliminating tt keeps the path but throws away the timing, which is why two different parametrizations can trace the same curve at different speeds or in opposite directions.

Example
Eliminate the parameter (polynomial): x=t2, y=tx=t^2,\ y=t
t = 2t = −2
Arrows show the direction of travel as tt increases: upward along x=y2x=y^2.

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:

t−2−1012x=t241014y=t−2−1012\begin{array}{c|ccccc} t & -2 & -1 & 0 & 1 & 2\\ \hline x=t^2 & 4 & 1 & 0 & 1 & 4\\ y=t & -2 & -1 & 0 & 1 & 2 \end{array}
Example
Eliminate the parameter (trig): x=3cos⁡t, y=2sin⁡tx=3\cos t,\ y=2\sin t

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

(x3)2+(y2)2=1⟹x29+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 0≤t≤10\le t\le1, use

x=1+(5−1)t=1+4t,y=2+(−4−2)t=2−6t.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 cos⁡2t+sin⁡2t=1\cos^2 t+\sin^2 t=1 (or sec⁡2t−tan⁡2t=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.

Try it
Quick check

Eliminate the parameter: x=2+tx=2+t, y=t2−1y=t^2-1. Describe the curve and its direction.

Show the answer

Answer: y=(x−2)2−1y=(x-2)^2-1, a parabola with vertex (2,−1)(2,-1), traced left to right as tt increases. From the first equation t=x−2t=x-2; substitute into the second.

Polar Equations of Conics (Introduction)

Concept
Eccentricity form

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

r=ed1±ecos⁡θorr=ed1±esin⁡θ,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.

  • e=1⇒e=1\Rightarrow parabola,   0<e<1⇒0<e<1\Rightarrow ellipse,   e>1⇒e>1\Rightarrow hyperbola.
  • cos⁡\cos means the directrix is vertical; sin⁡\sin means it is horizontal.
Example
Identify r=62+cos⁡θr=\dfrac{6}{2+\cos\theta}
focus (pole)
e=12e=\tfrac12: an ellipse with one focus at the pole, from r=2r=2 on the right to r=6r=6 on the left.

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

r=62+cos⁡θ=31+12cos⁡θ.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 cos⁡θ\cos\theta or sin⁡θ\sin\theta equal to ee.

Try it
Quick check

Identify r=82−2sin⁡θr=\dfrac{8}{2-2\sin\theta} and locate its directrix.

Show the answer

Answer: a parabola; directrix y=−4y=-4. Divide by 22: r=41−sin⁡θr=\dfrac{4}{1-\sin\theta}, so e=1e=1 and ed=4ed=4, giving d=4d=4. The −sin⁡θ-\sin\theta puts the directrix below the pole.

Where Conics Come From: Slicing a Cone

Concept
One cone, four curves
tilted slice → ellipse

Take a DOUBLE cone (two ice-cream cones tip to tip, extending forever) and cut it with a plane. The angle of the cut decides everything: a horizontal cut gives a circle; tilt the plane a little and the circle stretches into an ellipse.

Concept
The borderline cut: a parabola
slice parallel to the side → parabola

Tilt the plane until it is exactly PARALLEL to the cone's slant side and the curve never closes up — that single borderline angle produces the parabola. It is the knife-edge between the closed curves (ellipses) and the doubly-open one (hyperbola).

Concept
Cutting both nappes: a hyperbola
vertical slice (both nappes) → hyperbola

Tilt past the parallel position (or cut straight down) and the plane slices BOTH halves of the double cone, producing the two branches of a hyperbola. The two branches are one curve — the same plane meeting the same cone, twice.

Example
Why "conic sections" is literal

Circle, ellipse, parabola, hyperbola are not four unrelated graphs — they are the SAME construction (plane meets cone) at four tilts. That is why their equations are all degree-two: Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 covers every one of them.

Try it
Quick check

Match each eccentricity to its conic: e=0e=0, e=0.6e=0.6, e=1e=1, e=1.4e=1.4.

Show the answer

Answer: circle, ellipse, parabola, hyperbola. Eccentricity measures how far the cutting plane tilts from horizontal: 00 is the flat cut, 11 is the cut parallel to the side, and beyond 11 the plane catches both nappes.

The Ellipse

Concept
Anatomy of an ellipse

Standard form x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 (with a>ba > b): the long semi-axis is aa, the short one bb, and the two foci sit at (±c,0)(\pm c, 0) with c2=a2−b2c^2 = a^2 - b^2. The defining property: every point of the ellipse has distances to the two foci summing to 2a2a (a loop of string around two pins traces one).

Example
x²/25 + y²/9 = 1
(−4, 0)(4, 0)
Foci at (±4,0)(\pm4,0); from any point the two distances add to 2a=102a=10.

a=5a = 5, b=3b = 3, so c2=25−9=16c^2 = 25 - 9 = 16 and the foci are (±4,0)(\pm4, 0). Check the string property at (5,0)(5, 0): distances 11 and 99 sum to 10=2a10 = 2a ✓. Eccentricity e=ca=45e = \tfrac{c}{a} = \tfrac45 measures how stretched it is (e=0e = 0 is a circle).

Try it
Quick check

An ellipse has vertices (±5,0)(\pm5,0) and foci (±3,0)(\pm3,0). Write its equation.

Show the answer

Answer: x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1. a=5a=5 and c=3c=3, so b2=a2−c2=16b^2=a^2-c^2=16.

The Parabola

Concept
Focus and directrix

x2=4pyx^2 = 4py: focus (0,p)(0, p), directrix y=−py = -p, and every point of the curve is EQUIDISTANT from the focus and the directrix. Satellite dishes are parabolas because incoming parallel rays all bounce to the focus.

Example
y = x²/8
F(0, 2)y = −2
The point (4,2)(4,2) is 44 from the focus and 44 from the directrix.

Rewrite as x2=8yx^2 = 8y: 4p=84p = 8, so p=2p = 2 — focus (0,2)(0,2), directrix y=−2y = -2. The point (4,2)(4, 2): distance to focus =4= 4, distance to directrix =4= 4 ✓.

The Hyperbola

Concept
Asymptotes are the skeleton

x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 opens left-right with foci at (±c,0)(\pm c, 0), c2=a2+b2c^2 = a^2 + b^2 (note the PLUS — foci sit outside), and asymptotes y=±baxy = \pm\tfrac{b}{a}x: the branches hug these lines forever without touching. Defining property: the DIFFERENCE of the focal distances is constant (2a2a).

Example
x²/9 − y²/16 = 1
(5, 0)(−5, 0)
The 6×86\times8 box sets the asymptotes y=±43xy=\pm\tfrac43x; the foci sit at ±5\pm5.

a=3a = 3, b=4b = 4: c=5c = 5, foci (±5,0)(\pm5, 0), asymptotes y=±43xy = \pm\tfrac43x. At the vertex (3,0)(3,0) the focal distances are 22 and 88: difference 6=2a6 = 2a ✓.

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:

  • 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 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.
  • 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=40⟹c=±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=2x−210y=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=3⇒a2−b2=c2=9,tangency (m=1, c=5):c2=a2m2+b2⇒25=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 a2−b2=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):

  • 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 ℓ\ell (the semi-latus rectum) controls the polar form below.

parabola y2=4px: L.R.=∣4p∣;ellipse & hyperbola: L.R.=2b2a,ℓ=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 ℓ=a(1−e2)\ell=a(1-e^2) and for a hyperbola ℓ=a(e2−1)\ell=a(e^2-1), so knowing aa and ee recovers the full shape. Every focal chord of a conic has semi-latus rectum ℓ\ell as its harmonic-mean radius.

Concept
Classifying a general conic by B2−4ACB^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 B2−4ACB^2-4AC classifies the curve without ever finding the rotation angle:

  • B2−4AC<0⇒B^2-4AC<0\Rightarrow ellipse (circle if also A=C, B=0A=C,\ B=0),
  • B2−4AC=0⇒B^2-4AC=0\Rightarrow parabola,
  • B2−4AC>0⇒B^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
  • xy=1xy=1: here A=C=0, B=1A=C=0,\ B=1, so B2−4AC=1>0⇒B^2-4AC=1>0\Rightarrow hyperbola (rotated 45∘45^\circ).
  • 5x2−4xy+8y2−36=05x^2-4xy+8y^2-36=0: B2−4AC=(−4)2−4(5)(8)=16−160=−144<0⇒B^2-4AC=(-4)^2-4(5)(8)=16-160=-144<0\Rightarrow ellipse.
  • x2+23 xy+3y2−8x=0x^2+2\sqrt3\,xy+3y^2-8x=0: B2−4AC=(23)2−4(1)(3)=12−12=0⇒B^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+y−6=0x+y-6=0 and x+y−13=0x+y-\sqrt{13}=0 is

d=∣6−13∣12+12=6−132≈2.3941.414≈1.69.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=t−1tx=t+\dfrac{1}{t},\ y=t-\dfrac{1}{t} for t≠0t\ne0. Rather than solving for tt, combine the equations: x+y=2tx+y=2t and x−y=2tx-y=\dfrac{2}{t}, so their product removes tt:

(x+y)(x−y)=(2t) ⁣(2t)=4⟹x2−y2=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 ∣x∣≥2|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=asec⁡t, y=btan⁡tx=a\sec t,\ y=b\tan t uses sec⁡2t−tan⁡2t=1\sec^2t-\tan^2t=1 to yield x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.

Tip

Polar eccentricity form, revisited: in r=ℓ1±ecos⁡θr=\dfrac{\ell}{1\pm e\cos\theta} the numerator is the semi-latus rectum ℓ=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 ℓ\ell (the numerator), then recover a=ℓ1−e2a=\dfrac{\ell}{1-e^2} for an ellipse or a=ℓe2−1a=\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 B2−4ACB^2-4AC (not the AA-vs-CC test); and for rational or trig parametrizations always state the branch or orientation restriction the algebra hides.

Why It Works

Why does a parabolic mirror focus light? For y=14px2y = \tfrac{1}{4p}x^2 the tangent at (x0,y0)(x_0, y_0) makes equal angles with the vertical (incoming) ray and the line to the focus (0,p)(0, p) — a computation with the slope x02p\tfrac{x_0}{2p} and the reflection law. So every vertical ray reflects through the focus. The ellipse's foci have the analogous property, which is why whispering galleries work: sound from one focus reflects to the other.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Equation of a circle
(x−h)2+(y−k)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: (x−2)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 (x−h)2+(y−k)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.