$P_{1} : y^{2} = 4 a x, P_{2} : y^{2} = - 4 a x L : y = x$
If $a = 4,$ then equation of the ellipse having the line joining the focii of the parabolas $P_{1}$ and $P_{2}$ as the major axis and eccentricity equal to $\frac{1}{2}$ is

\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1

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\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1

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\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1

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\frac{x^{2}}{12} + \frac{y^{2}}{16} = 1

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Solution

The correct option is C $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$
$P_{1} : y^{2} = 4 a x$ and $P_{2} : y^{2} = - 4 a x .$
The focii of the parabolas are $(a, 0)$ and $(- a, 0)$ $= (4, 0), (- 4, 0)$
Major axis of ellipse $= 2 a = 2 \times 4$
Also,
eccentricity $e = \frac{1}{2}$
We know, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$\Rightarrow \frac{b^{2}}{a^{2}} = \frac{3}{4}$
$\Rightarrow b^{2} = 12$
And, also $a^{2} = 16$
Equation of the ellipse is: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ .
So, the correct option is (C).

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P1:y2=4ax,P2:y2=−4axL:y=xIf a=4, then equation of the ellipse having the line joining the focii of the parabolas P1 and P2 as the major axis and eccentricity equal to 12 is

$P_{1} : y^{2} = 4 a x, P_{2} : y^{2} = - 4 a x L : y = x$
If $a = 4,$ then equation of the ellipse having the line joining the focii of the parabolas $P_{1}$ and $P_{2}$ as the major axis and eccentricity equal to $\frac{1}{2}$ is