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(D) The equation of a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.Two para

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$2(a-b)$

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(D) The equation of a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.Two parallel normals correspond to parameters $	heta$ and $	heta + \pi$,which are $ax \sec 	heta - by \csc 	heta = a^2 - b^2$ and $ax \sec 	heta - by \csc 	heta = -(a^2 - b^2)$.The distance $L$ between these parallel lines is given by $L = \frac{|(a^2 - b^2) - (-(a^2 - b^2))|}{\sqrt{a^2 \sec^2 	heta + b^2 \csc^2 	heta}} = \frac{2(a^2 - b^2)}{\sqrt{a^2 \sec^2 	heta + b^2 \csc^2 	heta}}$.Using the inequality $a^2 \sec^2 	heta + b^2 \csc^2 	heta \ge (a+b)^2$,we get $L \le \frac{2(a^2 - b^2)}{(a+b)} = 2(a-b)$.Thus,the maximum value of $L$ is $2(a-b)$.

Let $L$ be the distance between two parallel normals of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Then the maximum value of $L$ is:

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