Q93 / 180 · KCET 2011 · Mathematics
If the focii of $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ and $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ coincide, then the value of $a$ is
- Correct answerA$\sqrt{3}$
- B$\frac{1}{\sqrt{3}}$
- C2
- D1
Solution
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
$\begin{aligned} b^{2} &=a^{2}\left(1-e^{2}\right) \\ 4 &=16\left(1-e^{2}\right) \\ \Rightarrow \quad e^{2} &=1-1 / 4=3 / 4 \\ e &=\pm \frac{\sqrt{3}}{4} \\ \text { Focii of an ellipse } &=(\pm a e, 0) \\ &=\left(\pm 4 \cdot \frac{\sqrt{3}}{2}, 0\right)=(\pm 2 \sqrt{3}, 0) \end{aligned}$
Given, focii of both conics are coincides.
$\Rightarrow \quad(\pm 2 \sqrt{3}, 0)=(\pm a e, 0)$
[ $\because$ Here $(\pm a e, 0)$ is focii of second conic.]
$\Rightarrow \quad \pm a e=\pm 2 \sqrt{3}$
$\Rightarrow \quad a^{2} e^{2}=12$
Equation of eccentricity of second conic (hyperbola)
$$
\begin{aligned}
&b^{2}=a^{2}\left(e^{2}-1\right) \\
&b^{2}=a^{2} e^{2}-a^{2}
\end{aligned}
$$
$$
\Rightarrow \quad b^{2}=a^{2} e^{2}-a^{2}
$$
$\begin{array}{ll}\Rightarrow & 9=12-a^{2} \\ \Rightarrow & a^{2}=3 \Rightarrow a=\sqrt{3}\end{array}$
Hence, the correct answer is option: 'A'.