Question

Paragraph :
Let $S$ and $S^{\prime }$ be the foci on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $P(x_1,y_1)$ be an arbitrary point as shown.
(d) Hence, find the relation between $\theta$ and $\varphi$ and verify the reflection property of hyperbola.

• A. $\theta =2\varphi$
• B. $2\theta =\varphi$
• C. $\theta =3\varphi$
• D. $\theta =\varphi$

Answer 1

The correct option is D $\theta =\varphi$

$tan\theta =|\frac{\frac{y_1}{x_1+ae}-\frac{b^2{x}_1}{a^2{y}_1}}{1+\left(\frac{y_1}{x_1+ae}\right)\left(\frac{b^2{x}_1}{a^2{y}_1}\right)}|$
$=|\frac{a^2{y}_1^2-b^2{x}_1^2-a{b}^2{x}_{1}e}{(a^2+b^2)x_1{y}_1+a^{3}e{y}_1}|$
$=|\frac{-a^2{b}^2-a{b}^2{x}_{1}e}{a^2{e}^2{x}_1{y}_1+a^{3}e{y}_1}|$
$=|\frac{-b^2(a{x}_{1}e+a^2)}{ae{y}_1(a{x}_{1}e+a^2)}|$
$=\frac{b^2}{ae{y}_1}$ since $y_1>0$ Hence, $\theta$ is acute.
Therefore, $tan\varphi =tan\theta$ and since $\theta$ and $\varphi$ are acute angles it follows that $\varphi =\theta$.
So, if to put the source of light into one of the two hyperbola's focus points and if the internal surface of the hyperbola reflects the light rays as a mirror, then all the light rays emitted by the source coincide after reflection with the straight rays released from the second hyperbola's focus point.
Hence option D is correct.