Question

A ray emanating from the point (5, 0) is incident on the hyperbola $9x^{2} - 16y^{2} = 144$ at the point P with abscissa 8; then the equation of reflected ray after first reflection is (Point P lies in first quadrant)

• A. $3\sqrt{3}x - 13y + 15\sqrt{3} = 0$
• B. $3x - 13y + 15 = 0$
• C. $3\sqrt{3}x + 13y - 15\sqrt{3} = 0$
• D. None of these

Answer 1

Answer: (A)

$3\sqrt{3}x - 13y + 15\sqrt{3} = 0$

Answer 2

Given hyperbola is $9x^{2} - 26y^{2} = 144$. This equation can be rewritten as $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ .....(i)

Since x coordinate of P is 8. Let y-coordinate of P is $\alpha$

$\therefore$ $(8,\alpha)$ lies on (i)

$\therefore$ $\frac{64}{16} - \frac{\alpha^{2}}{9} = 1$; $\therefore$ $\alpha = 27$ $(\because$P lies in first quadrant)

$\alpha = 3\sqrt{3}$

Hence coordinate of point P is $(8,3\sqrt{3})$

$\because$ Equation of reflected ray passing through $P(8,3\sqrt{3})$ and $S^{'}( - 5,0$); ∴ Its equation is $y - 3\sqrt{3} = \frac{0 - 3\sqrt{3}}{- 5 - 8}(x - 8)$

or $13y - 39\sqrt{3} = 3\sqrt{3}x - 24\sqrt{3}$ or $3\sqrt{3}x - 13y + 15\sqrt{3} = 0$