Question

A 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 on elliptical surface and get reflected as a beam having equation $kx-y+2(1-k)=0,\forall k\in R$. If locus of point of intersection of perpendicular tangents on ellipse is $x^2+y^2-4y-10=0$, then

• A. eccentricity of the ellipse is $\frac{1}{3}$
• B. eccentricity of the ellipse is $\frac{2}{3}$
• C. locus of foot of perpendicular from $(2,2)$ upon any tangent to the ellipse lies on curve $x^2+y^2-4y-5=0$
• D. locus of foot of perpendicular from $(2,2)$ upon any tangent to the ellipse lies on curve $x^2+y^2-4y-13=0$

Answer 1

Answer: (B), (C)

Equation of PS' is
$\lambda x-y+2+2\lambda =0$
$\Rightarrow \lambda (x+2)+(2-y)=0$
$\therefore x=-2,y=2$
focus $S^{\prime }=(-2,2)$
and PS is
$kx-y+2(1-k)=0$
$\Rightarrow k(x-2)+(2-y)=0$
$\therefore$ focus $S=(2,2)$
$\Rightarrow 2ae=4\Rightarrow ae=2$
$\because \text{ Director circle }$
$x^2+(y-2{)}^2=14$
$\therefore a^2+b^2=14\Rightarrow a^2+a^2\left(1-e^2\right)=14\Rightarrow a=3$
$\text{ and }b=\sqrt{5}$
$\therefore e=\frac{2}{3}$
$\therefore$ Locus of foot of perpendicular from focus is auxiliary circle
i.e. $x^2+(y-2{)}^2=3^2$
$\Rightarrow x^2+y^2-4y-5=0$