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 $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
$a^{2}=9, b^{2}=16 \quad$ 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.