NDA Maths · Conics

Hyperbola — Foci & Eccentricity

A hyperbola is the same family with e > 1; its standard form x²/a² − y²/b² = 1 has foci a distance c from the centre where c² = a² + b² — note the PLUS, the one sign that flips from the ellipse.

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Why this matters

Only 4 PYQs, all direct: read a and b from the standard form, use c² = a² + b² (not minus), and the foci and eccentricity follow. The parametric form (a sec θ, b tan θ) appears occasionally.

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Concept 1 of 2: Standard Form, Foci & Eccentricity

The hyperbola mirrors the ellipse with one change: the foci sit FURTHER out than the vertices, so c² = a² + b². Normalise the equation to '= 1', read a and b, and everything follows.

Definition

For $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ :

Foci: $(\pm c, 0)$ with $c^2 = a^2 + b^2$ ; distance between foci $= 2c$ .
Eccentricity: $e = \dfrac{c}{a} > 1$ .
Asymptotes: $y = \pm \dfrac{b}{a}x$ .
Always normalise to $=1$ first: $25x^2 - 75y^2 = 225$ becomes $\tfrac{x^2}{9} - \tfrac{y^2}{3} = 1$ .

Hyperbola foci & eccentricity

c^2 = a^2 + b^2, \qquad e = \dfrac{c}{a} > 1

Worked example

Find the distance between the foci of

\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1

The same idea in a real exam question:

NDA · Sep 2025 · Q65Easy

Example 1 · Conics · Hyperbola — Foci and Eccentricity

What is the distance between the foci of the hyperbola

x^2-4y^2=1

Hyperbola uses PLUS: c² = a² + b²

The ellipse has

c^2 = a^2 - b^2

; the hyperbola has

c^2 = a^2 + b^2

. Carrying the ellipse's minus sign into a hyperbola is the single most common slip in this subtopic.

A hyperbola's eccentricity is always greater than 1

Since

c^2 = a^2 + b^2 > a^2

, we have

c > a

and so

e = \dfrac{c}{a} > 1

for every hyperbola. An eccentricity computed as

<1

means you used the ellipse relation

c^2 = a^2 - b^2

by mistake.

Drill 1 more on standard form, foci & eccentricity

Concept 2 of 2: Parametric Form & θ-Independent Properties

A point given as (a sec θ, b tan θ) is on a hyperbola — recover the standard form using sec²θ − tan²θ = 1. Some hyperbola families have foci that don't move as a parameter changes, because c² stays constant.

Definition

Parametric point: $(a\sec\theta,\ b\tan\theta)$ lies on $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ (use $\sec^2\theta - \tan^2\theta = 1$ ). A point like $(3\tan\theta, 2\sec\theta)$ gives a hyperbola opening along $y$ .
$\theta$ -independent foci: for $\dfrac{x^2}{\cos^2\theta} - \dfrac{y^2}{\sin^2\theta} = 1$ , $c^2 = \cos^2\theta + \sin^2\theta = 1$ , so the foci are $(\pm 1, 0)$ regardless of $\theta$ , while $e = \sec\theta$ .
All three Pythagorean identities are in play here, not just one: $\sec^2\theta-\tan^2\theta=1$ (the parametric point), $\csc^2\theta-\cot^2\theta=1$ (a point $(a\csc\theta,\,b\cot\theta)$ is on the same hyperbola), and $\sin^2\theta+\cos^2\theta=1$ (an ellipse). Products like $\cot\theta\csc\theta$ collapse via $\csc^2-\cot^2=1$ when a chord length is asked.

Parametric identity

\sec^2\theta - \tan^2\theta = 1

Worked example

A point on a curve is

(2\sec\theta, \sqrt{5}\tan\theta)

. Find the eccentricity.

The same idea in a real exam question:

NDA · Apr 2021 · Q64Moderate

Example 2 · Conics · Hyperbola — Foci and Eccentricity

If any point on a hyperbola is

(3\tan\theta, 2\sec\theta)

, then what is the eccentricity of the hyperbola?

$(a\sec\theta, b\tan\theta)$ traces a HYPERBOLA, not an ellipse

The eliminating identity is

\sec^2\theta - \tan^2\theta = 1

, which gives

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

— a hyperbola. The ellipse uses

(a\cos\theta, b\sin\theta)

with

\cos^2\theta + \sin^2\theta = 1

. Confusing the two parametrisations flips the conic.

Drill 1 more on parametric form & θ-independent properties

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Standard Form, Foci & Eccentricity
Hyperbola foci & eccentricity
$c^2 = a^2 + b^2, \qquad e = \dfrac{c}{a} > 1$
Parametric Form & θ-Independent Properties
Parametric identity
$\sec^2\theta - \tan^2\theta = 1$

Watch out for (3)

Hyperbola uses PLUS: c² = a² + b²
→ Standard Form, Foci & Eccentricity
A hyperbola's eccentricity is always greater than 1
→ Standard Form, Foci & Eccentricity
$(a\sec\theta, b\tan\theta)$ traces a HYPERBOLA, not an ellipse
→ Parametric Form & θ-Independent Properties

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