Visual walkthrough — Parametric forms of all conics

Before anything, three plain-word warm-ups so no symbol arrives unannounced.

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Step 1 — Draw the one identity everything rests on

WHAT. Put a point $P$ on a circle of radius $1$ centred at the origin. Drop a straight line down to the horizontal axis. You now have a right triangle: a horizontal leg, a vertical leg, and the radius as the slanted side (the hypotenuse).

WHY. Every parametrization on this page is just this triangle, stretched. If we understand the circle's triangle, we understand all of them.

PICTURE. In the figure, the angle $\theta$ is measured at the centre, from the positive horizontal axis, turning anticlockwise (counter-clockwise). Read off the two legs:

$\underset{\text{horizontal leg}}{\underbrace{\cos \theta }},\underset{\text{vertical leg}}{\underbrace{\sin \theta }}.$

Pythagoras on this triangle (leg² + leg² = hypotenuse²) gives, since the hypotenuse is $1$:

$\underset{{\text{(horizontal)}}^2}{\underbrace{{\cos }^2\theta }}+\underset{{\text{(vertical)}}^2}{\underbrace{{\sin }^2\theta }}=\underset{{\text{radius}}^2}{\underbrace{1}}.$

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Step 2 — Force a circle to appear by copying that identity

WHAT. We want points that automatically satisfy $x^2+y^2=r^2$. We already have an equation whose right side is $1$. Multiply the whole identity by $r^2$:

$r^2{\cos }^2\theta +r^2{\sin }^2\theta =r^2.$

WHY. Now compare term by term with $x^2+y^2=r^2$. The shapes match exactly if we declare:

$x=\underset{\text{so }{x}^2=r^2{\cos }^2\theta }{\underbrace{r\cos \theta }},y=\underset{\text{so }{y}^2=r^2{\sin }^2\theta }{\underbrace{r\sin \theta }}.$

PICTURE. The triangle from Step 1 just got scaled up by $r$: horizontal leg becomes $r\cos \theta$, vertical leg becomes $r\sin \theta$, hypotenuse becomes $r$. Turning the dial $\theta$ walks $P$ once around the circle.

Recall Why is

$\theta$ genuinely the geometric angle here? For a circle the point's direction from the centre is exactly $\theta$ ::: because the legs $r\cos \theta ,r\sin \theta$ ARE the horizontal/vertical projections of the radius at angle $\theta$.

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Step 3 — Squash the circle into an ellipse

WHAT. Take every circle point $(a\cos \theta ,a\sin \theta )$ (radius $a$) and shrink only its height by the factor $b/a$. The vertical coordinate $a\sin \theta$ becomes $\frac{b}{a}\cdot a\sin \theta =b\sin \theta$.

WHY. A shrink in one direction turns a circle into an ellipse. We check it lands on the right curve by plugging into the ellipse equation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{(a\cos \theta {)}^2}{a^2}+\frac{(b\sin \theta {)}^2}{b^2}=\underset{=1}{\underbrace{{\cos }^2\theta +{\sin }^2\theta }}=1.✓$

The $a^2$ and $b^2$ in the denominators exactly cancel the $a^2$ and $b^2$ born from squaring — leaving the master identity behind.

PICTURE. The dashed circle of radius $a$ is the auxiliary circle. The blue vertical drop from a circle point to the squashed ellipse point shows the $b/a$ shrink. The angle $\theta$ lives on the circle, not on the line to the ellipse point — that is why we call it the eccentric angle, see Eccentric angle and auxiliary circle.

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Step 4 — The problem: a minus sign has no sin/cos solution

WHAT. The hyperbola equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Try the ellipse recipe: $x=a\cos \theta ,y=b\sin \theta$. Plug in:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}={\cos }^2\theta -{\sin }^2\theta .$

WHY it fails. ${\cos }^2\theta -{\sin }^2\theta$ is not always $1$ — it equals $\cos 2\theta$, which swings between $-1$ and $+1$. So these points do NOT stay on the hyperbola. We need a different identity whose left side is a difference of two squares equal to 1.

PICTURE. The figure plots the value ${\cos }^2\theta -{\sin }^2\theta$ as $\theta$ turns: a wave crossing $1$ only at isolated instants. Compare the flat green line at height $1$ (what we need) with the red wave (what sin/cos gives). They only touch here and there — hence the recipe breaks.

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Step 5 — Meet $\sec$ and $\tan$ on the same triangle

WHAT. Go back to the unit-circle right triangle of Step 1: legs $\cos \theta$ (horizontal), $\sin \theta$ (vertical), hypotenuse $1$. Define two new lengths:

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\text{vertical leg}}{\text{horizontal leg}},\sec \theta =\frac{1}{\cos \theta }=\frac{\text{hypotenuse}}{\text{horizontal leg}}.$

WHY these two. Divide the master identity ${\cos }^2\theta +{\sin }^2\theta =1$ through by ${\cos }^2\theta$:

$\frac{{\cos }^2\theta }{{\cos }^2\theta }+\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }⟹\underbrace{1}+\underset{(\sin /\cos {)}^2}{\underbrace{{\tan }^2\theta }}=\underset{(1/\cos {)}^2}{\underbrace{{\sec }^2\theta }}.$

Rearrange:

$\boxed{{\sec }^2\theta -{\tan }^2\theta =1}$

There it is — a difference of two squares that is always exactly $1$. This is the minus-sign identity we hunted for. See Trigonometric identities.

PICTURE. The figure draws the tangent line to the unit circle at the horizontal axis: $\tan \theta$ is the height where the extended radius meets that vertical tangent line; $\sec \theta$ is the length of that extended radius out to the meeting point. The right triangle formed (base $1$, height $\tan \theta$, hypotenuse $\sec \theta$) makes ${\sec }^2=1+{\tan }^2$ visible as Pythagoras again.

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Step 6 — Force the hyperbola to appear

WHAT. We now copy the structure of ${\sec }^2\theta -{\tan }^2\theta =1$ onto $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Declare:

$\frac{x}{a}=\sec \theta \Rightarrow x=a\sec \theta ,\frac{y}{b}=\tan \theta \Rightarrow y=b\tan \theta .$

WHY. Substitute and the denominators cancel the squared factors:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\underset{(x/a{)}^2}{\underbrace{{\sec }^2\theta }}-\underset{(y/b{)}^2}{\underbrace{{\tan }^2\theta }}=1.✓$

PICTURE. The figure traces the hyperbola as $\theta$ turns. For $\theta$ in $(-\frac{\pi }{2},\frac{\pi }{2})$, $\sec \theta >0$ → the right branch. As $\theta$ crosses into $(\frac{\pi }{2},\frac{3\pi }{2})$, $\sec \theta <0$ → the left branch. Colours mark which slice of $\theta$ paints which branch.

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Step 7 — All the cases: every branch, the forbidden angle, the alternative

WHAT. Walk through where $\theta$ may and may not go, and offer the branch-clean alternative.

• Right branch: $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, here $\cos \theta >0$ so $x=a\sec \theta \ge a>0$.
• Left branch: $\theta \in (\frac{\pi }{2},\frac{3\pi }{2})$, here $\cos \theta <0$ so $x=a\sec \theta \le -a<0$.
• Forbidden $\theta =\frac{\pi }{2}$ (and $\frac{3\pi }{2}$): $\cos \theta =0$, so $\sec \theta$ and $\tan \theta$ blow up to infinity. Geometrically these are the asymptotes — the point runs off to infinity along the slanted guide lines $y=\pm \frac{b}{a}x$. The parametrization should skip them; that is the curve genuinely having no point there.

WHY an alternative. With $\sec /\tan$ the single dial $\theta$ jumps between branches — awkward for calculus. The Hyperbolic functions give a branch-clean dial:

$\cosh t=\frac{e^t+e^{-t}}{2}\ge 1,\sinh t=\frac{e^t-e^{-t}}{2},{\cosh }^{2}t-{\sinh }^{2}t=1.$

Since $\cosh t\ge 1$ always, $x=a\cosh t\ge a>0$: this covers only the right branch, smoothly, for every real $t$. Use $x=-a\cosh t$ for the left branch.

PICTURE. Split figure: left half shows $\theta$ near $\frac{\pi }{2}$ sending the point out along the dashed asymptote; right half shows $t$ (real line) sweeping the right branch with no jump.

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The one-picture summary

Everything above is one triangle used three ways: kept (circle), squashed (ellipse), and extended to the tangent line (hyperbola). The figure stacks all three and labels which identity each uses.

Recall Feynman: tell the whole walk in plain words

Draw a right triangle inside a circle of radius 1. Its two legs are named $\cos \theta$ and $\sin \theta$, and Pythagoras says ${\cos }^2+{\sin }^2=1$. To make a circle of radius $r$, blow the triangle up by $r$. To make an ellipse, squash its height by $b/a$ — the same identity survives because the $a^2,b^2$ cancel. But a hyperbola has a minus, and sin/cos can only make a plus, so we hunt for a minus-identity. We find it by dividing the same Pythagoras equation by ${\cos }^2\theta$: that turns it into ${\sec }^2-{\tan }^2=1$, a difference that's always 1. Copy its shape and out drops $x=a\sec \theta ,y=b\tan \theta$. When $\cos \theta =0$ the point flies off to infinity — those are the asymptotes. For a smoother ride on one branch, swap in $\cosh ,\sinh$, which obey ${\cosh }^2-{\sinh }^2=1$ by the same difference-of-squares magic.

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Connections

• 3.4.12 Parametric forms of all conics (Hinglish) — the parent topic
• 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

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