Question

The equation(s) of an standard ellipse which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$, is/are

• A. $3x^2+5y^2=32$
• B. $3x^2+5y^2=48$
• C. $5x^2+3y^2=32$
• D. $5x^2+3y^2=48$

Answer 1

Answer: (A), (D)

$5x^2+3y^2=48$

Let ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
for horizontal ellipse $a>b$
Since, $(-3,1)$ lies on it
$\therefore \frac{9}{a^2}+\frac{1}{b^2}=1\cdots \left(1\right)$
for horizontal ellipse $a>b$
Given that $e=\sqrt{\frac{2}{5}}$
$\therefore b^2=a^2(1-e^2)$
$\Rightarrow 5b^2=3a^2\cdots \left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$, we get
$a^2=\frac{32}{3}$ and $b^2=\frac{32}{5}$
$\therefore$ Ellipse is
$\frac{x^2}{\left(\frac{32}{3}\right)}+\frac{y^2}{\left(\frac{32}{5}\right)}=1$
$\Rightarrow 3x^2+5y^2=32$

​​​​​​​for vertical ellipse $b>a$
Given that $e=\sqrt{\frac{2}{5}}$
$\therefore a^2=b^2(1-e^2)$
$\Rightarrow 5a^2=3b^2\cdots \left(3\right)$

from equation $\left(1\right)$ and $\left(3\right)$, we get
$a^2=\frac{48}{5}$ and $b^2=16$
$\therefore$ Ellipse is
$\frac{x^2}{\left(\frac{48}{5}\right)}+\frac{y^2}{16}=1$
$\Rightarrow 5x^2+3y^2=48$