Find the equation of the normal to the ellipse $9 x^{2} + 16 y^{2} = 288$ at the point $(4, 3) .$

4 x - 3 y = 7

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3 x - 4 y = 7

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4 x + 3 y = 7

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3 x + 4 y = 7

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Solution

The correct option is A $4 x - 3 y = 7$
Given ellipse may be written as, $\frac{x^{2}}{32} + \frac{y^{2}}{18} = 1$
$\Rightarrow a^{2} = 32, b^{2} = 18$
Hence required normal at $(4, 3)$ is given by,
$y - 3 = \frac{3 \times 32}{4 \times 18} (x - 4) \Rightarrow y - 3 = \frac{4}{3} (x - 4) \Rightarrow 4 x - 3 y = 7$

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