Question

If a hyperbola passes through the point $P(10,16)$ and it has vertices at $( ± 6,0)$, then the equation of the normal to it at P is :

• A. $3x + 4y=94$
• B. $x + 2y=42$
• C. $2x + 5y=100$
• D. $x + 3y=58$

Answer 1

Answer: (C)

$\frac{x^{2}}{36} - \frac{y^{2}}{b^{2}} \quad...\left(i\right)$
$P\left(10,16\right)$ lies on $\left(i\right)$ get $b^{2} = 144$
$\frac{x^{2}}{36} - \frac{y^{2}}{144} = 1$
Equation of normal is
$\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2}e^{2}$
$2x + 5y = 100$