Question

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

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

Answer 1

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

The correct options are
A The latus rectum of ellipse is $1.$
C Sum of the focal distances of a point $P(x,y)$ on the ellipse is $4.$
D $y=3$ meets the tangents drawn at the vertices of the ellipse at points $P$ and $Q$ then $PQ$ substends a right angle at any of its foci.

The given equation is: $x^2+{4y}^2-2x-16y+13=0$

$\Rightarrow x^2-2x+1+{4(y}^2-4y+4)-4=0$

$\Rightarrow {(x-1)}^2+4{(y-2)}^2=4$

$\Rightarrow \frac{{(x-1)}^2}{4}+\frac{{(y-2)}^2}{1}=1$

A] Length of latus rectum $=\frac{{2b}^2}{a}=1$

B] $e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{\frac{3}{4}}$

$\therefore$ Focal length $=2ae=2\times 2\times \frac{\sqrt{3}}{2}=2\sqrt{3}$

C] Sum of the focal distances $=2a=4$

D] $Y=y-2$ and $X=x-1$ then,

$Q(-2,1),P(2,1),S=(\sqrt{3},0)$

$m_1=\frac{0-1}{\sqrt{3}+2}=\frac{-1}{\sqrt{3}+2}$

$m_2=\frac{0-1}{\sqrt{3}-2}=\frac{-1}{\sqrt{3}-2}$

$\Rightarrow m_1\times m_2=\frac{1}{(\sqrt{3}+2)(\sqrt{3}-2)}$

$=\frac{1}{3-4}=-1$

$\therefore PQ$ subtends right angle at any one of its foci.

Options [A],[C],[D]