If lines 3x + 2y = 10 and -3x + 2y = 10 are t | vedclass

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Question

Let ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

tangent $\frac{\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{ye}}{\mathrm{b}}=1$

Vedclass · Accepted Answer

$\frac{{100}}{{27}}$

Vedclass · Answer

Let ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

tangent $\frac{\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{ye}}{\mathrm{b}}=1$

$	herefore \frac{b}{e}=5$

and $b=\frac{10}{3} \Rightarrow e=\frac{2}{3}$

$\Rightarrow \mathrm{a}^{2}=\frac{500}{81} \quad 	herefore \mathrm{L} \cdot \mathrm{R}=\frac{2 \mathrm{a}^{2}}{\mathrm{b}}=\frac{100}{27}$

If lines $3x + 2y = 10$ and $-3x + 2y = 10$ are tangents at the extremities of latus rectum of an ellipse whose centre is origin, then the length of latus rectum of ellipse is

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