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

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