Question

A set of lines $x+y-2+\lambda (2x+y-3)=0$ represents incident rays on an ellipse $S=0$ and
$2x+3y-23+\mu (2x-y-3)=0$ represents the set of reflected rays from the ellipse where $\lambda ,\mu \in R.$ If $P(3,7)$ is a point on the ellipse normal at which meets the major axis at $N$ then

• A. Eccentricity of ellipse is $\frac{\sqrt{5}}{2\sqrt{2}+1}$
• B. $N$ divides line segment joining two foci in the ratio $2\sqrt{2}:1$
• C. Area of triangle formed by point $P$ and two foci is $5$ Sq. unit.
• D. Eccentricity of ellipse is $\frac{\sqrt{5}}{2\sqrt{2}-1}$

Answer 1

Answer: (A), (B), (C)

Area of triangle formed by point $P$ and two foci is $5$ Sq. unit.

Since rays eminating from one focus after reflection passes through other focus.
One focus should be concurrency point of the first family of lines i.e. $S_1(1,1)$ and the other is point of concurrency of the second family i.e. $S_2(4,5)$.
Given $P(3,7)$ is a point on the ellipse
$\Rightarrow 2a=P{S}_1+P{S}_2\Rightarrow 2a=\sqrt{5}(2\sqrt{2}+1)$
And $2ae=S_1{S}_2=5$
$\therefore e=\frac{\sqrt{5}}{(2\sqrt{2}+1)}$
$\because$ Normal is angular bisector for triangle $S_{1}P{S}_2$
$\Rightarrow \frac{N{S}_1}{N{S}_2}=\frac{P{S}_1}{P{S}_2}=\frac{\sqrt{40}}{\sqrt{5}}=2\sqrt{2}$

Now for line $S_1{S}_2:4x-3y=1$
perpendicualr distance from $P$ is
$d=\frac{10}{5}=2$ unit
$\therefore$ Area of
$△P{S}_1{S}_2=\frac{1}{2}\cdot d\cdot S_1{S}_2=\frac{1}{2}\times 2\times 5=5\text{sq. unit}$