Question

Which of the following is/are true about the ellipse $x^2+4y^2-2x-16y+13=0$

• A. The length of latus rectum of the ellipse is $1$ unit
• B. Distance between foci of the ellipse is $4\sqrt{3}$ unit
• C. Sum of the focal distances of a point $P(x,y)$ on the ellipse is $4$ unit
• D. If the tangents drawn at the vertices of the ellipse meets line $y=3$ at the points $P$ and $Q$ then $PQ$ always subtends a right angle at any of its foci

Answer 1

Answer: (A), (C), (D)

If the tangents drawn at the vertices of the ellipse meets line $y=3$ at the points $P$ and $Q$ then $PQ$ always subtends a right angle at any of its foci

Given ellipse is $x^2+4y^2-2x-16y+13=0$
$\Rightarrow (x^2-2x+1)+4(y^2-4y+4)=4$
$\Rightarrow \frac{(x-1{)}^2}{4}+\frac{(y-2{)}^2}{1}=1$
$\therefore$ Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 1}{2}=1$ unit

Also $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

Distance between foci $=2ae$
$\Rightarrow 2ae=2\times 2\times \frac{\sqrt{3}}{2}=2\sqrt{3}$ unit

Sum of the focal distance $=2a=4$ unit

Tangents at the vertices are $x-1=\pm a$
$\Rightarrow x-1=\pm 2$
or $x=3,-1$
Therefore, the line $y=3$ intersect these at point $P(3,3)$
and $Q(-1,3)$.
Coordinates of focus are $S(\sqrt{3}+1,2)$
Slope of $PS$ is $\frac{1}{2-\sqrt{3}},$ slope of $QS$ is $\frac{1}{-2-\sqrt{3}}$
$\Rightarrow$ product of slopes $=\frac{1}{2-\sqrt{3}}\times \frac{1}{-2-\sqrt{3}}=-1$
It means $PS\perp QS$.