Q.The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$. Its equation is
(A) $x^2 - y^2 = 32$
(B) $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$
(C) $2x - 3y^2 = 7$
(D) none of these

Concept: Hyperbola – relationship between foci, eccentricity, and standard form.

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the distance between foci is $2c$ where $c^2 = a^2 + b^2$, and eccentricity $e = \frac{c}{a}$.

Step 1: From the given data, distance between foci $= 2c = 16$, so $c = 8$.

Step 2: Eccentricity $e = \sqrt{2} = \frac{c}{a} = \frac{8}{a}$, which gives $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$, hence $a^2 = 32$.

Step 3: Using $c^2 = a^2 + b^2$: $64 = 32 + b^2$, so $b^2 = 32$. …

The standard form of a hyperbola centered at the origin is either $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (horizontal transverse axis) or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (vertical transverse axis). The key parameters are:

• $a$: semi-transverse axis
• $b$: semi-conjugate axis
• $c$: distance from center to each focus, where $c^2 = a^2 + b^2$
• $e = \frac{c}{a}$: eccentricity

The distance between the two foci is $2c$, and we're told this equals 16. The eccentricity relates the focal distance to the size of the hyperbola.

1. Find the focal distance parameter $c$

Since the distance between foci is 16:

2. Use eccentricity to find $a$

The eccentricity is given as $e = \sqrt{2}$. Using $e = \frac{c}{a}$:

Therefore $a^2 = (4\sqrt{2})^2 = 32$.

3. Find $b^2$ using the fundamental relation

For a hyperbola, $c^2 = a^2 + b^2$:

4. Write the equation

Since $a^2 = b^2 = 32$, the equation is:

$\frac{x^2}{32} - \frac{y^2}{32} = 1$ …