What is the equation of the normal which is perpendicular to 3x + 4y = 5 for the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

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Solution

The correct option is D

We will first find the slope of the normal, m. We know the equation of normal with slope m.
$y = m x - \frac{(a^{2} - b^{2}) m}{\sqrt{a^{2} + b^{2} m^{2}}} S l o p e o f 3 x + 4 y = 5 = \frac{- 3}{4} S l o p e o f n o r m a l = \frac{- 1}{s l o p e o f 3 x + 4 y = 5} = \frac{\frac{- 1}{- 3}}{4} = \frac{4}{3}$
So the equation of normal becomes
$y = \frac{4}{3} x - \frac{(a^{2} - b^{2}) \times \frac{4}{3}}{\sqrt{a^{2} + b^{2} \times {(\frac{4}{3})}^{2}}} = \frac{4}{3} x - \frac{4}{3} \frac{\frac{(a^{2} - b^{2})}{\sqrt{9 a^{2} + 16 b^{2}}}}{3} y = \frac{4}{3} x - \frac{4 (a^{2} - b^{2})}{\sqrt{9 a^{2} + 16 b^{2}}}$

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