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  • If the focus of the parabola coincides with one of the foci of the ellipse <img alt="3 x^2+b y^2-12 x=0

If the focus of the parabola y^2-8=4 x coincides with one of the foci of the ellipse 3 x^2+b y^2-12 x=0, then the eccentricity of the ellipse is

Option: 1

\frac{1}{\sqrt{2}}


Option: 2

\frac{1}{4}


Option: 3

\frac{1}{2}


Option: 4

None of these 


Answers (1)

best_answer

The parabola is  y^2=4(x-2) and its focus is at (3,0).

The ellipse \frac{(x-2)^2}{4}+\frac{y}{\frac{12}{b}}=1 and its foci are at 

( \pm 2 e+2,0)

\Rightarrow \quad 3=2 e+2

Posted by

Riya

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