Parametric forms of all conics: Question bank

Two words we lean on constantly, so let's pin them down in plain language before any question:

Also note the letter $r$ : throughout, $r$ is the radius of the circle $x^{2} + y^{2} = r^{2}$ — a fixed positive number, the distance from the centre to every point of the circle.

The picture that unlocks the whole page is the auxiliary circle: the circle of radius $a$ drawn around the ellipse. The eccentric angle $θ$ is measured there, then the point is pushed straight down by the factor $b / a$ to land on the ellipse — which is exactly why the two angles disagree.

True or false — justify

True or false: For the circle

x = r cos θ, y = r sin θ

(where

r > 0

is the radius), the parameter

θ

is the actual geometric angle to the point.

True. The point sits at distance

r

in the direction

θ

, so its quadrant-aware direction

atan2 (r sin θ, r cos θ) = θ

— dial and protractor agree exactly, in every quadrant.

True or false: For the ellipse

x = a cos θ, y = b sin θ

θ

equals the geometric angle to the point.

False. The geometric direction is

atan2 (b sin θ, a cos θ)

. Because the ellipse's

y

-coordinate is the circle's

y

scaled by the factor

b / a

(a vertical affine scaling, not a rotation), the ratio

y / x

changes from

tan θ

\frac{b}{a} tan θ

— so the measured angle shifts even though

x

is untouched. Thus

θ

is the eccentric angle, not the protractor angle. See Eccentric angle and auxiliary circle and the figure above.

True or false:

(a cos θ, b sin θ)

can parametrize a hyperbola if we just pick

a, b

cleverly.

False. No choice of

a, b

makes a sum of squares into a difference; you'd need

cos^{2} - sin^{2}

which is not an identity. The minus sign forces

sec^{2} - tan^{2} = 1

instead.

True or false: As

t

runs over all reals,

(a t^{2}, 2 a t)

traces the entire parabola

y^{2} = 4 a x

True.

y = 2 a t

takes every real value once, and each fixes

x = a t^{2} \geq 0

uniquely — every point of the parabola is hit exactly once, no gaps.

True or false: In

x = a sec θ, y = b tan θ

, letting

θ

range over

[0, 2 π)

covers both branches of the hyperbola.

True.

sec θ \geq 1

(- \frac{π}{2}, \frac{π}{2})

(right branch) and

sec θ \leq - 1

(\frac{π}{2}, \frac{3 π}{2})

(left branch), so both are reached — provided

θ \neq = \frac{π}{2}, \frac{3 π}{2}

where

sec

blows up. The figure below shows how the point jumps to the left branch as

θ

crosses

\frac{π}{2}

True or false:

x = a cosh t, y = b sinh t

covers the whole hyperbola.

False. Since

cosh t \geq 1

always,

x \geq a > 0

, so only the right branch is traced. The left branch needs

x = - a cosh t

. See Hyperbolic functions.

True or false: A parametric form is valid only if substituting it back gives the Cartesian equation for some value of the parameter.

False. It must hold identically — for every value in the range. "For some value" would be satisfied by a single point, not the whole curve.

True or false: The circle is just the special case

a = b

of the ellipse parametrization.

True. Set

a = b = r

(r cos θ, r sin θ)

, and the auxiliary circle coincides with the ellipse, so the eccentric angle becomes the geometric angle again.

True or false: On a parabola

y^{2} = 4 a x

, two points

t_{1}

and

t_{2}

with

t_{2} = - t_{1}

are reflections across the axis.

True. They share

x = a t^{2}

but have

y = \pm 2 a t

, so they sit symmetrically above and below the

x

-axis — the axis of the parabola.

True or false: The parabola's tangent slope

\frac{1}{t}

works at every point, including the vertex

t = 0

False. At

t = 0

\frac{1}{t}

is undefined — because the tangent at the vertex

(0, 0)

is vertical (infinite slope).

\frac{d x}{d t} = 2 a t = 0

there, so the slope

\frac{d y / d t}{d x / d t}

divides by zero; the curve is momentarily going straight up. Every other

t

gives a finite, valid slope.

Spot the error

Find the error: "Ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

, so

x = 9 cos θ, y = 4 sin θ

You need

a = 9 = 3

and

b = 4 = 2

, not

9

and

4

. The parametrization uses

a, b

(semi-axis lengths), and the denominators are

a^{2}, b^{2}

— so

x = 3 cos θ, y = 2 sin θ

Find the error: "Parabola: point is

(t^{2}, 2 t)

for every parabola."

The constant

a

is missing. Correct is

(a t^{2}, 2 a t)

; check:

y^{2} = 4 a^{2} t^{2} = 4 a \cdot a t^{2} = 4 a x

. Dropping

a

only works for the specific case

a = 1

Find the error: "For hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

take

x = a tan θ, y = b sec θ

The roles are swapped. That gives

tan^{2} - sec^{2} = - 1

, not

+ 1

, so it satisfies the conjugate hyperbola

\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1

. You want

x = a sec θ, y = b tan θ

Find the error: "Tangent to

y^{2} = 4 a x

t

: differentiate to get

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 a t}{2 a} = t

The ratio is upside down: slope

= \frac{d y / d t}{d x / d t} = \frac{2 a}{2 a t} = \frac{1}{t}

, not

t

. Numerator is

d y / d t = 2 a

, denominator

d x / d t = 2 a t

— and this is undefined at

t = 0

(vertical tangent at the vertex). See Parametric differentiation.

Find the error: "Eliminate parameter from

x = a sec θ, y = b tan θ

: since

sec θ = x / a

and

tan θ = y / b

, we get

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

The identity is

sec^{2} θ - tan^{2} θ = 1

, so it must be a minus:

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

. Copying the ellipse's plus sign silently converts a hyperbola into an ellipse.

Find the error: "

θ = \frac{π}{2}

is fine for

x = a sec θ

; it just gives

x = 0

sec \frac{π}{2} = \frac{1}{c o s ( π /2 )} = \frac{1}{0}

is undefined, not

0

. That's why

θ = \frac{π}{2}

is excluded — it corresponds to the direction toward the asymptote where the point runs off to infinity.

Find the error: "For an ellipse, the point at

θ = \frac{π}{4}

lies on the line

y = x

from the centre."

Only if

a = b

. For

a \neq = b

the point is

(a / 2, b / 2)

, whose geometric direction is

atan2 (b / 2, a / 2) = arctan (b / a) \neq = 4 5^{\circ}

. Equal eccentric angle does not mean equal geometric angle.

Why questions

Why do we use a trig identity rather than solving the Cartesian equation for

y

Solving for

y

gives

y = \pm b 1 - x^{2} / a^{2}

— a two-valued, square-root mess with sign cases. A trig identity gives a single clean point for each

θ

and automatically stays on the curve for all

θ

Why does the ellipse use

sin / cos

but the hyperbola use

sec / tan

The ellipse equation is a sum of squares

= 1

, matched by

cos^{2} + sin^{2} = 1

; the hyperbola is a difference of squares

= 1

, matched by

sec^{2} - tan^{2} = 1

. The identity's sign structure must mirror the conic's.

Why is

y = 2 a t

(not

y = t

) the standard choice for the parabola?

Concretely compare the tangent lines. With

y = 2 a t

you get the point

(a t^{2}, 2 a t)

, slope

\frac{1}{t}

, and the tangent collapses to the clean

t y = x + a t^{2}

. With

y = t

you'd get the point

(\frac{t ^{2}}{4 a}, t)

, slope

\frac{2 a}{t}

, and the tangent

y t = 2 a x + \frac{t ^{2}}{2}

— carrying stray

4 a

's and halves. Both describe the same curve, but the

2 a

scaling is chosen precisely so the

4

y^{2} = 4 a x

cancels and the tangent formula has no leftover constants.

Why does parametric differentiation beat implicit differentiation for finding a tangent slope?

With one parameter,

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

is a plain quotient of ordinary derivatives — no need to solve an implicit relation between

x

and

y

or track

\pm

branch signs.

Why does the hyperbolic form

cosh / sinh

only give one branch while

sec / tan

gives both?

cosh t

never dips below

1

, so

x = a cosh t

stays

\geq a

(right branch only). But

sec θ

swings through both

\geq 1

and

\leq - 1

θ

crosses

\frac{π}{2}

, so

x

reaches both

+

and

-

sides.

Why does eliminating the parameter prove a parametric form is correct?

If you recover the Cartesian equation as an identity (true for all parameter values), you've shown every parametric point satisfies the curve's defining equation — which is exactly the definition of "lies on the curve."

Why is the eccentric angle defined via the auxiliary circle rather than the ellipse itself?

The auxiliary circle (radius

a

) gives an honest geometric angle; the ellipse point is that circle point pushed vertically by

b / a

. Borrowing the circle's angle keeps a clean geometric meaning even though it's no longer the direction to the ellipse point. See Eccentric angle and auxiliary circle.

Edge cases

What is the parametric point on

y^{2} = 4 a x

t = 0

(a \cdot 0, 2 a \cdot 0) = (0, 0)

— the vertex of the parabola. The parameter passes smoothly through the tip, but note its tangent there is vertical (slope

\frac{1}{t}

undefined), so it is a special point for slopes.

How does

tan θ

change sign as

θ

crosses

\frac{π}{2}

, and why does that matter for the hyperbola point?

Just below

\frac{π}{2}

tan θ \to + \infty

; just above,

tan θ \to - \infty

(it flips sign because

cos θ

flips from

0^{+}

0^{-}

). Meanwhile

sec θ

flips from

+ \infty

- \infty

, so the point

(a sec θ, b tan θ)

leaps from the far top-right of the right branch to the far bottom-left of the left branch. That sign flip is exactly how one continuous dial reaches both branches — see the s03 figure.

What happens to

(a sec θ, b tan θ)

θ \to \frac{π}{2}^{-}

Both

sec θ

and

tan θ \to + \infty

, and their ratio

\frac{t a n θ}{s e c θ} = sin θ \to 1

, so the point races off to infinity along the asymptote of slope

b / a

. This is the limiting behaviour, not an actual point.

Does

θ = 0

give a valid ellipse point, and where?

Yes:

(a cos 0, b sin 0) = (a, 0)

— the right vertex on the major axis. No degeneracy; it's a perfectly ordinary boundary point.

θ

and

θ + 2 π

give the same point on the ellipse

(a cos θ, b sin θ)

Yes —

cos

and

sin

are

2 π

-periodic, so

(a cos (θ + 2 π), b sin (θ + 2 π)) = (a cos θ, b sin θ)

. The ellipse is traced once per full turn, so its natural range is the half-open

[0, 2 π)

, exactly like the circle.

What conic does

x = a cos θ, y = b sin θ

become when

b \to 0

It collapses to the segment

y = 0, - a \leq x \leq a

traced back and forth — a degenerate ellipse (a line segment). The "curve" has zero height, so it's no longer a genuine 2D conic.

For the circle, do

θ

and

θ + 2 π

give the same point?

Yes —

cos, sin

are

2 π

-periodic, so the parameter wraps: the circle is traced once per full turn, which is why the range is written

[0, 2 π)

(half-open, to avoid repeating the start point).

Can two different parameter values on the same hyperbola give the same point in the

sec / tan

form?

Within one period

[0, 2 π) ∖ {\frac{π}{2}, \frac{3 π}{2}}

, no — each valid

θ

gives a distinct point. But

θ

and

θ + 2 π

repeat, exactly as with the circle, so the range must be restricted to one period.

x = a cosh t

ever equal to

x = - a

for real

t

No.

cosh t \geq 1

a cosh t \geq a > 0

always — confirming the left vertex

(- a, 0)

is unreachable by this form and belongs to the missing branch.

Recall One-line summary of every trap

Sum of squares → sin/cos → geometric angle only for the circle; difference of squares → sec/tan (both branches, via the sign flip of $tan$ at $\frac{π}{2}$ ) or cosh/sinh (right branch only); parabola is direct with the $a$ riding along (vertical tangent at the vertex $t = 0$ ); and a parametric form is right ⟺ eliminating the parameter reproduces the Cartesian equation identically.