Parametric forms of all conics: Worked examples

Before anything, one reminder in plain words. A parameter is a single dial (call it $θ$ or $t$ ) you turn; each dial-setting hands you one point $(x, y)$ that is guaranteed to sit on the curve. We never work with the messy Cartesian constraint again — we work with the dial.

The scenario matrix

Every problem you can meet lives in exactly one of these cells. The examples below are labelled by cell.

Cell	Case class	Covered by
A	Ellipse, first-quadrant point ( $θ$ acute)	E1
B	Ellipse, a point in each quadrant II/III/IV (sign bookkeeping)	E2
C	Degenerate: circle as $a = b$ , and the axis endpoints $θ = 0, \frac{π}{2}, π, \frac{3 π}{2}$	E3
D	Hyperbola, choosing the right vs left branch (sign of $sec θ$ )	E4
E	Hyperbola, the forbidden value $θ = \frac{π}{2}$ and the asymptote limit	E5
F	Parabola, a point above and below the axis (sign of $t$ ), and vertex $t = 0$	E6
G	Limiting case: a chord becomes a tangent as two parameters merge	E7
H	Word problem: a satellite / real orbit (ellipse)	E8
I	Exam twist: eliminate the parameter when it is disguised	E9

The building blocks (earned before use)

The sign of $sec θ$ = the sign of $cos θ$ . So $sec θ > 0$ (angle near $0$ ) gives $x = a sec θ > 0$ → right branch; $sec θ < 0$ (angle near $π$ ) gives $x < 0$ → left branch. Hold that thought for E4.

See Trigonometric identities for the identities $cos^{2} + sin^{2} = 1$ and $sec^{2} - tan^{2} = 1$ we lean on, and Hyperbolic functions for the $cosh / sinh$ alternative.

Worked examples

Worked example E2 — One point in

each of quadrants II, III, IV · Cell B Same ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ . Find the points at $θ = \frac{2 π}{3}$ (QII), $θ = \frac{5 π}{4}$ (QIII), $θ = \frac{7 π}{4}$ (QIV).

Forecast: which coordinate is negative in each? Predict the sign pattern before computing.

$θ = \frac{2 π}{3}$ : $cos = - \frac{1}{2}, sin = + \frac{3}{2}$ → $x = 4 (- \frac{1}{2}) = - 2, y = 2 \cdot \frac{3}{2} = 3$ . Left & up = QII. ✓ Why this step? The sign of $cos$ and $sin$ carries the quadrant — the $a, b$ only scale, they never flip signs.
$θ = \frac{5 π}{4}$ : $cos = - \frac{2}{2}, sin = - \frac{2}{2}$ → $x = 4 (- \frac{2}{2}) = - 22, y = 2 (- \frac{2}{2}) = - 2$ . Both $-$ = QIII. ✓
$θ = \frac{7 π}{4}$ : $cos = + \frac{2}{2}, sin = - \frac{2}{2}$ → $x = 22, y = - 2$ . Right & down = QIV. ✓ Why? Reading the unit-circle picture (figure s01) tells you the sign in each region without memorising.

Verify (QIII point): $\frac{( - 2 2 ) ^{2}}{16} + \frac{( - 2 ) ^{2}}{4} = \frac{8}{16} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1.$ ✓ (Signs vanish under squaring — that is why the same identity holds in every quadrant.)

Worked example E3 — Degenerate cases: circle & axis endpoints ·

Cell C (a) Show $\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1$ has $θ$ = the true geometric angle. (b) List the four axis-endpoints of $\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1$ .

Forecast: for part (a), when does the eccentric angle equal the real angle? (Guess: when the ellipse is a circle.)

(a) Here $a^{2} = b^{2} = 9 \Rightarrow a = b = 3$ , so $x = 3 cos θ, y = 3 sin θ$ — the circle form. Why this step? When $a = b$ there is no vertical squash, so the eccentric angle and the geometric angle coincide. This is the ONLY case the parent's mistake-callout allows them to be equal.
(b) $a = 5, b = 2$ . Endpoints occur at the "clean" angles:
- $θ = 0$ : $(5, 0)$ — right vertex.
- $θ = \frac{π}{2}$ : $(0, 2)$ — top co-vertex.
- $θ = π$ : $(- 5, 0)$ — left vertex.
- $θ = \frac{3 π}{2}$ : $(0, - 2)$ — bottom co-vertex. Why? At these four angles one of $cos, sin$ is $0$ and the other is $\pm 1$ , so a coordinate collapses to $0$ — that is a degenerate (zero) input and it lands you exactly on an axis.

Verify: $(0, 2)$ : $\frac{0}{25} + \frac{4}{4} = 0 + 1 = 1.$ ✓ $(- 5, 0)$ : $\frac{25}{25} + 0 = 1.$ ✓

Worked example E4 — Hyperbola: choosing the branch by sign of

$sec θ$ · Cell D On $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ find (a) a point with $θ = \frac{π}{3}$ , (b) a point with $θ = \frac{2 π}{3}$ , and say which branch each is on.

Forecast: predict the sign of $x$ from the sign of $cos θ$ before computing.

$a = 3, b = 4$ (from $a^{2} = 9, b^{2} = 16$ ). Form: $x = 3 sec θ, y = 4 tan θ$ .
(a) $θ = \frac{π}{3}$ : $cos = \frac{1}{2} > 0$ , so $sec = 2$ → $x = 3 \cdot 2 = 6 > 0$ . $tan \frac{π}{3} = 3$ → $y = 43$ . Since $x > 0$ : right branch. Why this step? $x = a sec θ \geq a$ whenever $cos θ > 0$ ; the point cannot enter the gap $(- 3, 3)$ , exactly matching the hyperbola.
(b) $θ = \frac{2 π}{3}$ : $cos = - \frac{1}{2}$ , so $sec = - 2$ → $x = 3 \cdot (- 2) = - 6 < 0$ . $tan \frac{2 π}{3} = - 3$ → $y = - 43$ . Since $x < 0$ : left branch. Why? One trig parametrization reaches both branches — that is the advantage over the $cosh$ form, which (with a $+$ sign) only gives the right branch.

Verify (a): $\frac{36}{9} - \frac{( 4 3 ) ^{2}}{16} = 4 - \frac{48}{16} = 4 - 3 = 1.$ ✓ (b): $\frac{36}{9} - \frac{48}{16} = 4 - 3 = 1.$ ✓

Worked example E5 — The forbidden

$θ = \frac{π}{2}$ and the asymptote limit · Cell E Why is $θ = \frac{π}{2}$ banned in $x = a sec θ$ ? And what happens to $(a sec θ, b tan θ)$ as $θ \to \frac{π}{2}^{-}$ ?

Forecast: guess where the point runs off to — a corner? infinity along a straight line?

At $θ = \frac{π}{2}$ , $cos θ = 0$ , so $sec θ = \frac{1}{0}$ is undefined. No point exists there. Why this step? Division by zero — the same reason $tan \frac{π}{2}$ blows up. This is the "degenerate input" cell for the hyperbola.
As $θ \to \frac{π}{2}^{-}$ : $cos θ \to 0^{+}$ , so $x = a sec θ \to + \infty$ and $y = b tan θ \to + \infty$ .
Ratio: $\frac{y}{x} = \frac{b tan θ}{a sec θ} = \frac{b}{a} \cdot \frac{sin θ / cos θ}{1/ cos θ} = \frac{b}{a} sin θ \to \frac{b}{a} \cdot 1 = \frac{b}{a}$ . Why this step? We formed the slope from the origin to see the direction the point escapes. It tends to $\frac{b}{a}$ — the slope of the asymptote $y = \frac{b}{a} x$ .
So the missing $θ = \frac{π}{2}$ corresponds to "the point at infinity along the asymptote." The gap in the parameter is the asymptote.

Verify: take $θ = 1.5$ rad (just under $\frac{π}{2} \approx 1.5708$ ) on the E4 hyperbola: slope $\frac{b}{a} sin θ = \frac{4}{3} sin 1.5 \approx 1.330$ , close to $\frac{4}{3} \approx 1.333$ . ✓ (See Tangents and normals to conics for how asymptotes emerge as limiting tangents.)

Worked example E7 — Limiting case: chord → tangent on the parabola ·

Cell G On $y^{2} = 8 x$ ( $a = 2$ ) write the chord through parameters $t_{1}$ and $t_{2}$ , then let $t_{2} \to t_{1} = t$ and recover the tangent.

Forecast: the parent found the tangent slope is $1/ t$ . Guess: the chord slope should be $\frac{2}{t _{1} + t _{2}}$ , which becomes $\frac{2}{2 t} = \frac{1}{t}$ .

Points $P = (2 t_{1}^{2}, 4 t_{1})$ , $Q = (2 t_{2}^{2}, 4 t_{2})$ . Slope $= \frac{4 t _{1} - 4 t _{2}}{2 t _{1}^{2} - 2 t _{2}^{2}} = \frac{4 ( t _{1} - t _{2} )}{2 ( t _{1} - t _{2} ) ( t _{1} + t _{2} )} = \frac{2}{t _{1} + t _{2}}$ . Why this step? We cancelled the common factor $(t_{1} - t_{2})$ — legal while $t_{1} \neq = t_{2}$ , i.e. while it is a genuine chord.
Let $t_{2} \to t_{1} = t$ : slope $\to \frac{2}{t + t} = \frac{1}{t}$ . Why? A tangent is the limit of chords as the two touch-points merge. This is where the derivative comes from — see Parametric differentiation.
Line through $(2 t^{2}, 4 t)$ with slope $\frac{1}{t}$ : $y - 4 t = \frac{1}{t} (x - 2 t^{2}) \Rightarrow t y - 4 t^{2} = x - 2 t^{2} \Rightarrow t y = x + 2 t^{2}$ . Why? This is the parent's $t y = x + a t^{2}$ with $a = 2$ — consistency check passed.

Verify: parent formula gives $t y = x + a t^{2} = x + 2 t^{2}$ ; our derivation gives the same. At $t = 2$ , tangent $2 y = x + 8$ ; touch-point $(8, 8)$ satisfies $2 \cdot 8 = 8 + 8 = 16$ . ✓

Recall

Recall Sign bookkeeping — the one thing to never forget

In the ellipse form, which factor decides the quadrant? ::: The signs of $cos θ$ and $sin θ$ — the $a, b$ only scale. On the hyperbola $x = a sec θ$ , right branch corresponds to which sign of $cos θ$ ? ::: $cos θ > 0$ (so $sec θ > 0$ , $x > a > 0$ ). Which single $θ$ is forbidden for $x = a sec θ$ and what does it "become"? ::: $θ = \frac{π}{2}$ ; it is the point at infinity along the asymptote. On the parabola, why do $t$ and $- t$ share an $x$ but mirror $y$ ? ::: $x = a t^{2}$ is even in $t$ ; $y = 2 a t$ is odd in $t$ . A chord becomes a tangent when… ::: the two parameters merge ( $t_{2} \to t_{1}$ ) — the limit of chords.

Connections

3.4.12 Parametric forms of all conics (Hinglish)
Conic Sections - Standard equations
Eccentric angle and auxiliary circle
Tangents and normals to conics
Chord of contact and pole-polar
Trigonometric identities
Hyperbolic functions
Parametric differentiation