Question

A coplanar beam of light emerging from a point source have the equation $\lambda x–y+2(1+\lambda )=0,\forall \lambda \in R$; the rays of the beam strike an elliptical surface and get reflected inside the ellipse. The reflected rays form another convergent beam having the equation $\mu x-y+2(1-\mu )=0,\forall \mu \in R$. Further it is found that the foot of the perpendicular from the point $(2,2)$ upon any tangent to the ellipse lies on the circle $x^2+y^2–4y–5=0$.

• A. The eccentricity of the ellipse is equal to $\frac{2}{3}$.
• B. The eccentricity of the ellipse is equal to $\frac{1}{3}$
• C. The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to $2\sqrt{5}$.
• D. The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to $4\sqrt{5}$

Answer 1

Answer: (A), (C)

The correct options are
A The eccentricity of the ellipse is equal to $\frac{2}{3}$.
C The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to $2\sqrt{5}$.

$(2-y)+\lambda (x+2)=0\Rightarrow$ family of lines through $(-2,2)$
And $(2,-y)+\mu (x-2)=0\Rightarrow$ family of lines tharough $(2,2)$
and $x^2+y^2-4y-5=0$ will be auxilary circle & its radius $=a=3$
Distance between foci $=2ac=4$
$\Rightarrow ae=2$
$\therefore e=\frac{2}{3}$

Area of $\Delta P{F}_1{F}_2=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 4\times \text{height}$
maximum area = maximum height = $b$

$\Rightarrow b^2=a^2(1-e^2)=9(1-\frac{4}{9})$
$\Rightarrow b=\sqrt{5}$
$\text{Area}=2\sqrt{5}$