Question

The focus and corresponding directrix of an ellipse are $\left(3,4\right)$ and $x+y-1=0$ respectively. If the eccentricity of the ellipse is $\frac{1}{2}$ , then the coordinates of the centre of the ellipse are

• A. $\left(2,3\right)$
• B. $\left(4,5\right)$
• C. $\left(8,9\right)$
• D. $\left(1,2\right)$

Answer 1

Answer: (B)

Let, the lengths of the semi-major axis and semi-minor axis of the ellipse are $a$ & $b$ respectively, then the distance of the focus $\left(3,4\right)$ from the corresponding directrix $\left(x + y - 1 = 0\right)$ is equal to $\frac{a}{e}-ae$
$\Rightarrow \frac{a}{\frac{1}{2}}-a\left(\frac{1}{2}\right)=\left|\frac{3 + 4 - 1}{\sqrt{2}}\right|=3\sqrt{2}$
$\Rightarrow \frac{3 a}{2}=3\sqrt{2}\Rightarrow a=2\sqrt{2}$
Distance between the focus and centre $=ae=\sqrt{2}$
The slope of the axis of the ellipse is $1=tan \theta$
$\Rightarrow \left(cos \theta , sin ⁡ \theta \right)=\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ or $\left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}}\right)$
The points on the axis of the ellipse at a distance $\sqrt{2}$ units from (3,4) are $\left(3 \pm \sqrt{2}\left(\frac{1}{\sqrt{2}}\right), 4 \pm \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\right)$
$\Rightarrow$ Points are $\left(2,3\right)$ or $\left(4,5\right)$ , but the centre is $\left(4,5\right)$ because it is far from the directrix as compared to $\left(2,3\right)$ .