Let the foci and length of the latus rectum o | vedclass

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Question

$	ext { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} $

$ae=5 \quad b^2=\frac{5 \sqrt{2} a}{2} $

$b^2=a^2\left(1-e^2ight)=\frac{5 \sqr

Vedclass · Accepted Answer

$51$

Vedclass · Answer

$	ext { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} $

$ae=5 \quad b^2=\frac{5 \sqrt{2} a}{2} $

$b^2=a^2\left(1-e^2ight)=\frac{5 \sqrt{2} a}{2}$

$ \Rightarrow \mathrm{a}\left(1-\mathrm{e}^2ight)=\frac{5 \sqrt{2}}{2}$

$\Rightarrow \frac{5}{\mathrm{e}}\left(1-\mathrm{e}^2ight)=\frac{5 \sqrt{2}}{2} $

$\Rightarrow \sqrt{2}-\sqrt{2} \mathrm{e}^2=\mathrm{e} $

$ \Rightarrow \sqrt{2} \mathrm{e}^2+\mathrm{e}-\sqrt{2}=0 $

$ \Rightarrow \sqrt{2} \mathrm{e}^2+2 \mathrm{e}-\mathrm{e}-\sqrt{2}=0$

$ \Rightarrow \sqrt{2} \mathrm{e}(\mathrm{e}+\sqrt{2})-1(1+\sqrt{2})=0 $

$\Rightarrow(\mathrm{e}+\sqrt{2})(\sqrt{2} \mathrm{e}-1)=0 $

$ 	herefore \mathrm{e} 
eq-\sqrt{2} $

$\mathrm{e}=\frac{1}{\sqrt{2}}$

$\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1 \quad a=5 \sqrt{2} $

$ b=5$

$ a^2 b^2=b^2\left(e_1^2-1ight) \Rightarrow e_1^2= 51$

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