Question

A ray emanating from the point $(4,0)$ is incident on the ellipse $9x^2+25y^2=225$ at the point $P$ with abscissa $3$. Find the equation of the reflected ray after first reflection.

• A. $12x+5y=48$
• B. $12x-5y=48$
• C. $-12x-35y=48$
• D. $-12x+35y=48$

Answer 1

Answer: (C), (D)

The correct options are
C $-12x-35y=48$
D $-12x+35y=48$

Solution:

Given equation of ellipse: $9x^2+25y^2=225$

$\frac{x^2}{25}+\frac{y^2}{9}=1$

Let $P(3,y_1)$

or, $9\times 3^2+25y_1^2=225$

or, $25y_1^2=144$

or, $y_1=\pm \frac{12}{5}$

$\frac{x^2}{25}+\frac{y^2}{9}=1$

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

or, $e^2=1-\frac{b^2}{a^2}$

or, $e^2=1-\frac{9}{25}=\frac{16}{25}$

or, $e=\frac{4}{5}$

$ae=5\times \frac{4}{5}=4$

It means the given point $(4,0)$ is focus of the ellipse.

We know that rays emanating from the one focus passes through other focus i.e $(-4,0)$

So, Equation of reflected ray $\Rightarrow y=\frac{\frac{12}{5}-0}{3+4}(x+4)$

or, $35y=12x+48$

or, $-12x+35y=48$

and another equation of reflected ray$\Rightarrow y=\frac{\frac{-12}{5}-0}{3+4}(x+4)$

or, $35y=-12x-48$

or, $-12x-35y=48$