Find the equation of hyperbola given foci and eccentricity

Question

Find the equation of the hyperbola whose foci are at $(\pm 5, 0)$ and eccentricity is $\frac{5}{3}$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

Since the foci are at $(\pm 5, 0)$ — on the x-axis, symmetric about the origin — the hyperbola has the standard form:

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

For this form, foci are at $(\pm c, 0)$ where $c^2 = a^2 + b^2$ .

Foci at $(\pm 5, 0)$ gives us:

c = 5

Eccentricity $e = c/a$ , so:

\frac{5}{3} = \frac{5}{a}

a = 3 \implies a^2 = 9

c^2 = a^2 + b^2

25 = 9 + b^2

b^2 = 16

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

Why This Works

A hyperbola is defined as the locus of points where the difference of distances from two fixed points (foci) is constant ( $= 2a$ ). The eccentricity $e > 1$ tells us how “spread out” the hyperbola is — larger $e$ means the branches open wider.

The relationship $c^2 = a^2 + b^2$ for a hyperbola is analogous to $c^2 = a^2 - b^2$ for an ellipse. The key difference: for an ellipse, $c < a$ (foci inside), while for a hyperbola, $c > a$ (foci outside the vertices).

The three parameters $a$ , $b$ , $c$ are connected by one equation, so we need two independent pieces of information (here: foci location giving $c$ , and eccentricity giving the $c/a$ ratio).

Alternative Method — Using the definition directly

If given the foci and $e$ , you can also find the directrix: $x = \pm a/e = \pm 3/(5/3) = \pm 9/5$ .

Then use the focus-directrix property: for any point on the hyperbola, $\frac{\text{distance from focus}}{\text{distance from directrix}} = e$ .

This method is longer but useful when the hyperbola is not centred at the origin.

For JEE, memorise these key relations: Ellipse: $c^2 = a^2 - b^2$ , $e < 1$ . Hyperbola: $c^2 = a^2 + b^2$ , $e > 1$ . The sign difference in the $c^2$ formula is the most commonly confused point. A quick check: for a hyperbola, $c > a$ always (the foci are farther from the centre than the vertices).

Common Mistake

Students sometimes use $c^2 = a^2 - b^2$ (the ellipse formula) for a hyperbola. This gives a negative $b^2$ , which makes no sense. For a hyperbola, it’s always $c^2 = a^2 + b^2$ . Another error: if foci are on the y-axis, the standard form becomes $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ — students sometimes use the x-form regardless of foci orientation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the definition directly

Common Mistake

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