Question

Equation of the ellipse whose centre at $(0,0)$, major axis on the $y$-axis and passes through the points $(3,2)$ and $(1,6)$ is:

• A. $\frac{x^2}{40}+\frac{y^2}{10}=1$
• B. $\frac{x^2}{10}+\frac{y^2}{40}=1$
• C. $\frac{x^2}{10}-\frac{y^2}{40}=1$
• D. None of the above

Answer 1

Answer: (B)

$\frac{x^2}{10}+\frac{y^2}{40}=1$

Major axis is y-axis
Let the ellipse be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, (3, 2) and (1, 6) lies on it
$\therefore \frac{9}{b^2}+\frac{4}{a^2}=1$ .......... (i);
$\frac{1}{b^2}+\frac{36}{a^2}=1$ ........... (ii)
Subtracting $\frac{9-1}{b^2}+\frac{4-36}{a^2}=0\Rightarrow \frac{8}{a^2}-\frac{32}{a^2}=0$
$\Rightarrow 8a^2=32b^2$ $\therefore a^2=4b^2$
Putting the value, it becomes $\frac{9}{b^2}+\frac{4}{4b^2}=1$
$\therefore \frac{10}{b^2}=1$
$\therefore b^2=10$
Now, $a^2=4b^2=4\times 10=40$
$\therefore$ Equation of the ellipse is $\frac{x^2}{10}+\frac{y^2}{40}=1$