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  • If the point of intersection of the ellipse  <img alt="\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{a}^2\left(1-\mathrm{e}^2\right)}=1" src="https://entrancecorner.oncode

If the point of intersection of the ellipse  \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{a}^2\left(1-\mathrm{e}^2\right)}=1 and the hyperbola   \mathrm {\frac{x^2}{a^2}-\frac{y^2}{a^2\left(E^2-1\right)}=\frac{1}{4}}  side of \mathrm{y}-axis then

Option: 1

E=\frac{\sqrt{\mathrm{e}^2+24}-3 \mathrm{e}}{2}


Option: 2

\mathrm{E}=\frac{\sqrt{\mathrm{e}^2+5}-\mathrm{c}}{2}


Option: 3

\mathrm{E}=\frac{\sqrt{\mathrm{e}^2+1}-\mathrm{e}}{2}


Option: 4

\mathrm{E}=\frac{\sqrt{\mathrm{e}^2+12}-3 \mathrm{e}}{2}


Answers (1)

best_answer

Let the point of intersection be \mathrm{(x, y)} then  \mathrm{a-e x=E x-(a / 2)}. Also\mathrm{x=\frac{a e+\frac{a}{2} E}{2}} .

Eliminating \mathrm{x} we get,

\mathrm{E^2+3 e E+\left(2 e^2-6\right)=0 \Rightarrow \mathrm{E}=\frac{\sqrt{\mathrm{e}^2+24}-3 \mathrm{e}}{2}}

Posted by

Sanket Gandhi

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