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$\mathrm{P \equiv(2 a, a)}$ is a point on the ellipse $\mathrm{\frac{x^2}{24}+\frac{y^2}{12}=1}$ . If $\mathrm{S}$ and $\mathrm{S^{\prime}}$ are the foci of the ellipse find the equation of the external bisector of $\mathrm{\angle S P S '.}$
Option: 1
$\mathrm{x-y=3}$

Option: 2
$\mathrm{x-y=6}$

Option: 3
$\mathrm{x+y=6}$

Option: 4
$\mathrm{2 x+y=6}$

Answers (1)

best_answer

$\mathrm{\text { Let } P \equiv(2 a, a) \text { lies on } \frac{x^2}{24}+\frac{y^2}{12}=1}$

$\mathrm{\Rightarrow \frac{a^2}{6}+\frac{a^2}{12}=1 \Rightarrow P \equiv(4,2)}$

By property tangent at P externally bisects $\mathrm{ \angle S^{\prime} P S \Rightarrow}$ Equation of the external bisector $=$ Equation of tangent at $\mathrm{ P \equiv(4,2)}$

$\mathrm{ \Rightarrow \frac{4 x}{24}+\frac{2 y}{12}=1 \Rightarrow x+y=6 }$

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Pankaj Sanodiya

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