Question

If a ray of light incident along the line $x=5$ gets reflected from the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ at point $P$, then

• A. equation reflected line is $9x+40y+45=0$
• B. equation reflected line is $9x-40y+45=0$
• C. area of $△PS{S}^{\prime }=22.5$ sq. unit
  $\text{(}S,S^{\prime }\text{are foci of hyperbola)}$
• D. area of $△PS{S}^{\prime }=11.25$ sq. unit
  $\text{(}S,S^{\prime }\text{are foci of hyperbola)}$

Answer 1

Answer: (A), (B), (D)

area of $△PS{S}^{\prime }=11.25$ sq. unit

$\text{(}S,S^{\prime }\text{are foci of hyperbola)}$
The given hyperbola is
$\frac{x^2}{16}-\frac{y^2}{9}=1$
$\Rightarrow e=\frac{5}{4}$
Its foci are $(\pm 5,0)$
$\because$ given line passes through $(5,0)$
Hence $P(5,\pm \frac{9}{4})$
So the reflected ray must pass through $(-5,0)$ and $P$
Hence the required equation is
$y=\frac{\pm 9/4}{(5+5)}(x+5)\Rightarrow 40y=\pm 9(x+5)$
​​​​Area of triangle bounded between the incident line, reflected line and transverse axis is
$=\frac{1}{2}\times 2.25\times 10=11.25$ sq. unit