Exercises — Parametric forms of all conics

WHAT: we must match $x^2+y^2=49$. WHY the identity: ${\cos }^2\theta +{\sin }^2\theta =1$ is a sum of squares — same shape as $x^2+y^2$. Multiply the identity by the radius squared, i.e. by $49=7^2$: $49{\cos }^2\theta +49{\sin }^2\theta =49.$ So $x=7\cos \theta ,y=7\sin \theta$. Option (b) fails because $(49\cos \theta {)}^2=2401{\cos }^2\theta$, giving radius $49$ not $7$. Option (c) is a hyperbola form. Answer: (a).

• Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ → sum of squares → ${\cos }^2\theta +{\sin }^2\theta =1$.
• Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ → difference of squares → ${\sec }^2\theta -{\tan }^2\theta =1$ (or ${\cosh }^{2}t-{\sinh }^{2}t=1$).
• Parabola $y^2=4ax$ → no sum/difference identity; it directly links $y^2$ to $x$, so we set $y=2at$ and read off $x=a{t}^2$.

Here $a^2=16\Rightarrow a=4$ and $b^2=4\Rightarrow b=2$. The form is $x=a\cos \theta ,y=b\sin \theta$. $x=4\cos \frac{\pi }{4}=4\cdot \frac{\sqrt{2}}{2}=2\sqrt{2},y=2\sin \frac{\pi }{4}=2\cdot \frac{\sqrt{2}}{2}=\sqrt{2}.$ Check (definition of "lies on curve"): $\frac{(2\sqrt{2}{)}^2}{16}+\frac{(\sqrt{2}{)}^2}{4}=\frac{8}{16}+\frac{2}{4}=\frac{1}{2}+\frac{1}{2}=1.$ ✓ Point: $(2\sqrt{2},\sqrt{2})$.

Compare $y^2=12x$ with the standard $y^2=4ax$: $4a=12\Rightarrow a=3$. The point $t$ is $(a{t}^2,2at)=(3t^2,6t)$. At $t=2$: $(3\cdot 4,6\cdot 2)=(12,12)$. Check: $y^2=12^2=144$ and $12x=12\cdot 12=144$. ✓

$a=3,b=4$, form $x=a\sec \theta ,y=b\tan \theta$. $\sec \frac{\pi }{3}=\frac{1}{\cos (\pi /3)}=\frac{1}{1/2}=2$, and $\tan \frac{\pi }{3}=\sqrt{3}$. $x=3\cdot 2=6,y=4\sqrt{3}.$ Check: $\frac{36}{9}-\frac{(4\sqrt{3}{)}^2}{16}=4-\frac{48}{16}=4-3=1.$ ✓

WHAT: solve each equation for the trig function, then use the identity that connects them. $\sec \theta =\frac{x}{5},\tan \theta =\frac{y}{3}.$ WHY ${\sec }^2-{\tan }^2$: because that is the identity these two functions satisfy — ${\sec }^2\theta -{\tan }^2\theta =1$. ${\left(\frac{x}{5}\right)}^2-{\left(\frac{y}{3}\right)}^2=1⟹\frac{x^2}{25}-\frac{y^2}{9}=1.$ A hyperbola with $a=5,b=3$. Eliminating the parameter proves the form was right — it reverses the derivation.

The geometric angle $\varphi$ from the centre to the point satisfies $\tan \varphi =\frac{y}{x}=\frac{b\sin \theta }{a\cos \theta }=\frac{b}{a}\tan \theta .$

• If $a=b$ (a circle), $\tan \varphi =\tan \theta \Rightarrow \varphi =\theta$: the parameter is the geometric angle.
• If $a\ne b$, then $\tan \varphi =\frac{b}{a}\tan \theta \ne \tan \theta$ in general, so $\varphi \ne \theta$. What it looks like (figure): start on the big auxiliary circle of radius $a$ at true angle $\theta$ (lavender point), then squash vertically by factor $b/a$ to land on the ellipse (coral point). Squashing bends the radius line, so the coral point's angle $\varphi$ is smaller than $\theta$ in the first quadrant. That is why $\theta$ is the eccentric angle — see Eccentric angle and auxiliary circle — not $\arctan (y/x)$.

$\cosh t=\frac{e^t+e^{-t}}{2}$ is always $\ge 1$ (sum of a positive number and its reciprocal, minimised at $t=0$ giving $1$). So $x=a\cosh t\ge a>0$ for all $t$: only the right branch. Covering the other branch: use $x=-a\cosh t$. So the full hyperbola needs $x=\pm a\cosh t,y=b\sinh t$. The trig form $a\sec \theta$ does reach both branches, because $\sec \theta$ swings through $\pm \infty$ as $\theta$ crosses $\frac{\pi }{2}$ and becomes negative in quadrant II/III.

Here $4a=8\Rightarrow a=2$, so point $t$ is $(a{t}^2,2at)=(2t^2,4t)$. At $t=1$: $(2,4)$. Slope by Parametric differentiation: $\frac{dx}{dt}=4at=4t$ wait — carefully, $x=2t^2\Rightarrow \frac{dx}{dt}=4t$, and $y=4t\Rightarrow \frac{dy}{dt}=4$. Slope $=\frac{dy/dt}{dx/dt}=\frac{4}{4t}=\frac{1}{t}$. At $t=1$, slope $=1$. Tangent line through $(2,4)$: $y-4=1\cdot (x-2)\Rightarrow y=x+2$. Cross-check with the general formula $ty=x+a{t}^2$: with $t=1,a=2$: $y=x+2$. ✓ (see Tangents and normals to conics)

$P=(a{\alpha }^2,2a\alpha ),Q=(a{\beta }^2,2a\beta )$. $\text{slope}=\frac{2a\alpha -2a\beta }{a{\alpha }^2-a{\beta }^2}=\frac{2a(\alpha -\beta )}{a({\alpha }^2-{\beta }^2)}=\frac{2a(\alpha -\beta )}{a(\alpha -\beta )(\alpha +\beta )}=\frac{2}{\alpha +\beta }.$ WHY factor ${\alpha }^2-{\beta }^2$: it is a difference of squares $=(\alpha -\beta )(\alpha +\beta )$; the $(\alpha -\beta )$ cancels the numerator's, leaving a clean symmetric answer. Sanity limit: let $\beta \to \alpha$ (chord becomes tangent): slope $\to \frac{2}{2\alpha }=\frac{1}{\alpha }$, matching the tangent slope $1/t$ from Q9. ✓

Here $a=1$. Substitute $\alpha =1,\beta =3$: $y(1+3)=2x+2\cdot 1\cdot (1\cdot 3)\Rightarrow 4y=2x+6\Rightarrow 2y=x+3.$ Check the endpoints: point $t=1$ is $(1,2)$: $2\cdot 2=4$ and $1+3=4$ ✓. Point $t=3$ is $(9,6)$: $2\cdot 6=12$ and $9+3=12$ ✓.

Setting $\beta \to \alpha$: $\frac{\alpha +\beta }{2}\to \alpha$ and $\cos \frac{\alpha -\beta }{2}\to \cos 0=1$. So the tangent at angle $\alpha$ is $\frac{x}{a}\cos \alpha +\frac{y}{b}\sin \alpha =1.$ At $\alpha =\frac{\pi }{2}$: $\cos \frac{\pi }{2}=0,\sin \frac{\pi }{2}=1$, giving $\frac{y}{b}=1\Rightarrow y=b$. What it means: the tangent at the top of the ellipse is the horizontal line $y=b$ — exactly the height of the top co-vertex, as geometry demands. ✓ (This limit-of-a-chord idea is the engine behind Chord of contact and pole-polar.)

By Parametric differentiation: $\frac{dx}{d\theta }=a(1-\cos \theta ),\frac{dy}{d\theta }=a\sin \theta ,$ $\frac{dy}{dx}=\frac{a\sin \theta }{a(1-\cos \theta )}=\frac{\sin \theta }{1-\cos \theta }.$ At $\theta =\frac{\pi }{2}$: $\sin \frac{\pi }{2}=1,\cos \frac{\pi }{2}=0$, so $\frac{dy}{dx}=\frac{1}{1-0}=1$. Degenerate check: as $\theta \to 0$, $1-\cos \theta \to 0$ and the slope $\to \infty$ — the cycloid has a vertical cusp at $\theta =0$, matching the picture of a wheel-point touching the ground. All cases accounted for.

Validity: $xy=ct\cdot \frac{c}{t}=c^2$ for every $t\ne 0$ (the $t$ cancels), so the point always lies on the curve. The exclusion $t\ne 0$ is essential: $t=0$ would need $\frac{c}{t}$, undefined — and indeed the curve never touches the axes. Tangent via parametric differentiation: $\frac{dx}{dt}=c,\frac{dy}{dt}=-\frac{c}{t^2}$, so slope $=\frac{-c/t^2}{c}=-\frac{1}{t^2}$. Line through $(ct,\frac{c}{t})$: $y-\frac{c}{t}=-\frac{1}{t^2}(x-ct)\Rightarrow t^{2}y-ct=-x+ct\Rightarrow x+t^{2}y=2ct.$ Check the point lies on it: $ct+t^2\cdot \frac{c}{t}=ct+ct=2ct$ ✓.

Ellipse point at eccentric angle $\theta$ ::: $(a\cos \theta ,b\sin \theta )$ Hyperbola trig point at $\theta$ ::: $(a\sec \theta ,b\tan \theta )$ Parabola $y^2=4ax$ point $t$ ::: $(a{t}^2,2at)$ Parabola chord slope through $\alpha ,\beta$ ::: $\frac{2}{\alpha +\beta }$ Tangent to $y^2=4ax$ at $t$ ::: $ty=x+a{t}^2$ Tangent to $xy=c^2$ at $t$ ::: $x+t^{2}y=2ct$ Which branch does $(a\cosh t,b\sinh t)$ give? ::: the right branch only ($\cosh t\ge 1$)