Question

The equation of the normal to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at the point $(8,3\sqrt{3})$ is

• A. $\sqrt{3}x + 2y = 25$
• B. $x + y = 25$
• C. $y + 2x = 25$
• D. $2x + \sqrt{3}y = 25$

Answer 1

Answer: (D)

$2x + \sqrt{3}y = 25$

Answer 2

From $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$

Here $a^{2} = 16$, $b^{2} = 9$ and $(x_{1},y_{1}) = (8,3\sqrt{3})$

⇒ $\frac{16x}{8} + \frac{9y}{3\sqrt{3}} = 16 + 9$ i.e., $2x + \sqrt{3}y = 25$