Question

The eccentricities of the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1,\alpha >\beta$; and $\frac{x^2}{9}+\frac{y^2}{16}=1$ are equal. Which one of the following is correct?

• A. $4\alpha =3\beta$
• B. $\alpha \beta =12$
• C. $4\beta =3\alpha$
• D. $9\alpha =16\beta$

Answer 1

Answer: (A)

Let eccentricity of both the parabolas be $e$.
Then in the given ellipse: $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$
We have $a^2={\alpha }^2,b^2={\beta }^2$
$b^2=a^2\left(1-e^2\right)$
$\Rightarrow {\beta }^2={\alpha }^2\left(1-e^2\right)(\because \alpha >\beta )$
$\Rightarrow \frac{{\beta }^2}{{\alpha }^2}=1-e^2$
$\Rightarrow e^2=1-\frac{{\beta }^2}{{\alpha }^2}...(i)$
From equation $\frac{x^2}{9}+\frac{y^2}{16}=1$
$a^2=9,b^2=16$ Then $b^2=a^2\left(1-e^2\right),b>a$
$\frac{16}{9}=1-e^2$
$e^2=1-\frac{16}{9}...(ii)$
From equations (i) and (ii) we get:
$1-\frac{16}{9}=1-\frac{{\beta }^2}{{\alpha }^2}$
$\Rightarrow \frac{16}{9}=\frac{{\beta }^2}{{\alpha }^2}$
$\Rightarrow \frac{\beta }{\alpha }=\pm \frac{4}{3}$
$\Rightarrow 4\alpha =3\beta$
or $4\alpha =-3\beta$;
$4\alpha =3\beta$ is in the option.