Question

If major axis is the x-axis and passes through the points $(4,3)$ and $(6,2)$, then the equation for the ellipse whose centre is the origin is satisfies the given condition.

• A. $\frac{x^2}{52}+\frac{y^2}{13}=1$
• B. $\frac{x^2}{13}+\frac{y^2}{52}=1$
• C. $\frac{x^2}{52}-\frac{y^2}{13}=1$
• D. $\frac{x^2}{13}-\frac{y^2}{52}=0$

Answer 1

The correct option is A $\frac{x^2}{52}+\frac{y^2}{13}=1$

Major axis is x - axis
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$(4,3)$ and $(6,2)$ lies on it

$\frac{16}{a^2}+\frac{9}{b^2}=1$ ........ $\left(i\right)$

$\frac{36}{a^2}+\frac{4}{b^2}=1$ ........ $\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$, we get

$\frac{-20}{a^2}+\frac{5}{b^2}=0$

$\Rightarrow 5a^2=20b^2\Rightarrow a^2=4b^2$
Putting the value of $a^2$ in eqn. $\left(i\right)$

$\frac{16}{4b^2}+\frac{9}{b^2}=1$

$\therefore b^2=13$ and $a^2=4b^2=4\times 13=52$
$\therefore$ Equation of ellipse is $\frac{x^2}{52}+\frac{y^2}{13}=1$