Question

If an ellipse having its center at $(1,-1)$ ,length of semi-major axis as $8$ units passes through the point $(1,3)$, then

• A. eccentricity will be $\frac{\sqrt{3}}{2}$
• B. area of the quadrilateral whose diagonals are axes of ellipse is $64$ sq. units
• C. equation of directrices are $x=\pm \frac{16}{\sqrt{3}}$
• D. equation of directrices are $x=\pm \frac{16}{\sqrt{3}}+1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A eccentricity will be $\frac{\sqrt{3}}{2}$
B area of the quadrilateral whose diagonals are axes of ellipse is $64$ sq. units
D equation of directrices are $x=\pm \frac{16}{\sqrt{3}}+1$

Let the equation of the ellipse, centred at $(1,-1),$ be
$\Rightarrow \frac{(x-1{)}^2}{a^2}+\frac{(y+1{)}^2}{b^2}=1\dots \left(i\right)$
case$\left(i\right)$: If major axis is parallel to $Y$ axis i.e., $b>a$
then $b=8$
It passes through point $(1,3)$.
$\Rightarrow \frac{(1-1{)}^2}{a^2}+\frac{(3+1{)}^2}{64}=1$
$\Rightarrow 4=1$ ,which is not possible

Case$\left(ii\right)$: If major axis is parallel to $X$ axis i.e., $a>b$
then $a=8$
It passes through point $(1,3)$.
$\Rightarrow \frac{(1-1{)}^2}{64}+\frac{(3+1{)}^2}{b^2}=1$
$\Rightarrow \frac{16}{b^2}=1$
$\Rightarrow b^2=16$

Substituting the values of $a$ and $b$ in equation $\left(i\right)$, we get $\frac{(x-1{)}^2}{64}+\frac{(y+1{)}^2}{16}=1$

eccentricity of the ellipse will be
$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$
directrix equation will be
$x-1=\frac{a}{e}\Rightarrow x=\pm \frac{16}{\sqrt{3}}+1$
area of the quadrilateral with axes as diagonal is
$\frac{1}{2}\left(2a\right)\left(2b\right)=64$ sq. unit