Question
If straight line $\frac{ax}{3} + \frac{by}{4} = c$ is a normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$, then $a^{2} - b^{2}$ is equal to:
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Solution
Sol.(b) Let $\frac{ax}{3} + \frac{by}{4} = c$ is normal to ellipse at $(acos\theta,bsin\theta)$ Equation of normal to ellipse at $(acos\theta,bsin\theta)$ is
$$\begin{array}{r} axsec\theta - bycosec\theta = a^{2} - b^{2}\#(1) \end{array}$$
$$\begin{array}{r} \frac{ax}{3} + \frac{by}{4} = c\#(2) \end{array}$$
(1) and (2) are identical
$$\begin{matrix} & \ \Rightarrow \ 3sec\theta = - 4cosec\theta = \frac{a^{2} - b^{2}}{c} \\ & \ \Rightarrow \ cos\theta = \frac{3c}{a^{2} - b^{2}},sin\theta = \frac{- 4c}{a^{2} - b^{2}} \\ & \ \Rightarrow \ \left( \frac{3c}{a^{2} - b^{2}} \right)^{2} + \left( \frac{- 4c}{a^{2} - b^{2}} \right)^{2} = 1 \\ & \ \Rightarrow \ a^{2} - b^{2} = 5c \end{matrix}$$
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