Question

For hyperbola $\frac{x^2}{16}-\frac{y^2}{25}=1$, the equations of the directrices are

• A. $x=\frac{\pm 16}{\sqrt{41}}$
• B. $x=\frac{\pm 41}{\sqrt{16}}$
• C. $x=\frac{\pm 6}{\sqrt{41}}$
• D. $x=\frac{\pm 25}{\sqrt{41}}$

Answer 1

The correct option is A $x=\frac{\pm 16}{\sqrt{41}}$

Given equation is $\frac{x^2}{16}-\frac{y^2}{25}=1$

Here $a^2=16,b^2=25$

Equation of directrix of hyperbola is $\frac{a^2}{\sqrt{a^2+b^2}}$ and $\frac{-a^2}{\sqrt{a^2+b^2}}$

$=\frac{16}{\sqrt{16+25}}$ and $\frac{-16}{\sqrt{16+25}}$

So, answer is option A.