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 :

Updated On: Feb 14, 2025
  • $3x + 4y=94$
  • $x + 2y=42$
  • $2x + 5y=100$
  • $x + 3y=58$
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The Correct Option is C

Solution and Explanation

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