Question

If the focii of $\frac {x^2}{16}+\frac{y^2}{4}=1 $ and $\frac {x^2}{a^2}-\frac{y^2}{3}=1 $ coincide, then value of $a$ is

• A. $1$
• B. $\frac {1} {\sqrt {3}}$
• C. $2$
• D. $\sqrt {3}$

Answer 1

The Correct Option is D

Equation of conics
$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1,\,\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$
Equation of eccentricity of an ellipse
$b^{2} =a^{2}\left(1-e^{2}\right)$
$4 =16\left(1-e^{2}\right)$
$\Rightarrow e^{2} =1-1 / 4=3 / 4$
$e =\pm \frac{\sqrt{3}}{4}$
Focii of an ellipse $=(\pm a e, 0)$
$=\left(\pm 4 \cdot \frac{\sqrt{3}}{2}, 0\right)=(\pm 2 \sqrt{3}, 0)$
Given, focii of both conics are coincides.
$\Rightarrow (\pm 2 \sqrt{3}, 0)=(\pm a e, 0)$
$[\because$ Here $(\pm a e, 0)$ is focii of second conic.]
$\Rightarrow \pm a e=\pm 2 \sqrt{3}$
$\Rightarrow a^{2} e^{2}=12$
Equation of eccentricity of second conic (hyperbola)
$b^{2}=a^{2}\left(e^{2}-1\right)$
$\Rightarrow b^{2}=a^{2} e^{2}-a^{2}$
$\Rightarrow 9=12-a^{2}$
$\Rightarrow a^{2}=3$
$\Rightarrow a=\sqrt{3}$