यदि OB, एक दीर्घवृत्त का अर्ध लघुअक्ष है, F _ | vedclass

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Question

Let $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be he equation of ellipse.

Given that  ${F_1}B$ and ${F_2}B$ a

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

Let $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be he equation of ellipse.

Given that  ${F_1}B$ and ${F_2}B$ are perpandicluar to each other.

Slope of ${F_1}B 	imes $ alop of ${F_2}B =  - 1$

$\left( {\frac{{0 - b}}{{ - ae - 0}}} ight) 	imes \left( {\frac{{0 - b}}{{ae - 0}}} ight) =  - 1$

$\left( {\frac{b}{{ae}}} ight) 	imes \left( {\frac{{ - b}}{{ae}}} ight) =  - 1$

${b^2} = {a^2}{e^2}$

${e^2} = \frac{{{b^2}}}{{{a^2}}}\,\,\,$         {$\because $ $\,\,{e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}\,\,$}

$1 - \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{{{b^2}}}{{{a^2}}}\,$

$1 = 2\frac{{{b^2}}}{{{a^2}}}\,\,\, \Rightarrow \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{1}{2}$

${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}\,\,\, = 1 - \frac{1}{2} = \frac{1}{2}$

${e^2} = \frac{1}{2}$

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