C is the centre of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and L is an end of a latus rectum. If the normal at L meets the major axis in G, then CG =

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Solution

The correct option is C
Equation of normal at $L (a e, \frac{b^{2}}{a})$ is $\frac{a x}{e} - a y = a^{2} - b^{2}$
Since it meets the major axis at G then G $= (\frac{e}{a} (a^{2} - b^{2}), 0) = (- a e^{3}, 0) .$ Hence $C G = a e^{3}$

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\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

e^{4} + e^{2} =

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

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e^{4} + e^{2} - 1 = 0

e^{2} = \frac{\sqrt{5 - 1}}{2}

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