Question

For hyperbola $-\frac{x^2}{16}+\frac{y^2}{25}=1$ equation of directrices are

• A. $y=\pm \frac{16}{\sqrt{41}}$
• B. $y=\pm \frac{5}{\sqrt{41}}$
• C. $y=\pm \frac{2}{\sqrt{41}}$
• D. $y=\pm \frac{25}{\sqrt{41}}$

Answer 1

Answer: (D)

$y=\pm \frac{25}{\sqrt{41}}$

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

Directrix is at $y=\pm \frac{b}{e}$ where $e=\sqrt{\frac{b^2}{a^2}+1}$

Here $a^2=25,b^2=16⟹e=\sqrt{\frac{16}{25}+1}=\sqrt{\frac{41}{25}}$

So equation of directrix is $y=\pm \frac{5\sqrt{25}}{\sqrt{41}}=\pm \frac{25}{\sqrt{41}}$