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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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