Question

From the focus (-5, 0) of the ellipse $\frac{x^2}{45} + \frac{y^2}{20} = 1$, a ray of light is sent which makes an angle $cos^{-1}(\frac{-1}{\sqrt{5}})$ with the positive direction of $x$-axis, upon reaching the ellipse surface, the ray is reflected from it. Slope of the reflected ray is

• A. $-\frac{3}{2}$
• B. $-\frac{7}{3}$
• C. $-\frac{5}{4}$
• D. $-\frac{2}{11}$

Answer 1

Answer: (D)

The slope of the reflected ray is $-\frac{2}{11}$.

Answer 2

An important property of an ellipse is that a ray emanating from one focus reflects off the ellipse to the other focus. For the ellipse

$\frac{x^2}{45}+\frac{y^2}{20}=1$,

we have $a^2=45$ and $b^2=20$ so that

$c^2=a^2-b^2=45-20=25$, $c=5$.

Thus the foci are at $(-5,0)$ and $(5,0)$. The given ray originates from the focus $(-5,0)$. By the reflection property of ellipses, the reflected ray will pass through the other focus $(5,0)$.

The point of incidence on the ellipse is determined by the ray starting at $(-5,0)$ making an angle $\cos^{-1}\left(\frac{-1}{\sqrt{5}}\right)$ with the positive $x$-axis. Its initial direction vector is

$\mathbf{v}=\left(-\frac{1}{\sqrt5},\frac{2}{\sqrt5}\right)$.

Parameterizing the ray from $(-5,0)$:

$x=-5-\frac{t}{\sqrt5}$, $y=\frac{2t}{\sqrt5}$.

Substituting into the ellipse equation and solving, we find the intersection at $t=\sqrt5$, which yields the point

$P=(-5-1,\,0+2) = (-6,2)$.

Since the ellipse’s reflection property guarantees that the reflected ray from $P$ goes to the other focus $(5,0)$, the reflected ray has direction

$\mathbf{r} = (5 - (-6),\,0-2) = (11,-2)$.

Thus, the slope of the reflected ray is

$m=\frac{-2}{11}$.