Question

If the tangent drawn at the point $P(3\sqrt{2}, 4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ meets its directrix at $Q(\alpha, \beta)$ in the fourth quadrant then $\beta = $

• A. $\frac{5\sqrt{2}-9}{4}$
• B. $-\frac{9}{5}$
• C. $\frac{12\sqrt{2}-20}{5}$
• D. $-\frac{5}{4}$

Hint:

A key property of conics is that the tangent and the normal at any point P bisect the angle between the focal radii to that point. Another property is related to the tangent and directrix, which is used here. Always check if the given point lies on the conic before proceeding.

Answer 1

The Correct Option is C

Step 1: Equation of Tangent: Hyperbola: \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \). Point \( P(3\sqrt{2}, 4) \). Tangent at P: \( \frac{x(3\sqrt{2})}{9} - \frac{y(4)}{16} = 1 \implies \frac{\sqrt{2}x}{3} - \frac{y}{4} = 1 \).
Step 2: Find Directrix: Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} \). Directrices are \( x = \pm \frac{a}{e} = \pm \frac{3}{5/3} = \pm \frac{9}{5} \). Since Q is in the 4th quadrant (\( x>0, y<0 \)), we choose the positive directrix \( x = \frac{9}{5} \). So \( \alpha = \frac{9}{5} \).
Step 3: Find \( \beta \) (y-coordinate): Substitute \( x = \frac{9}{5} \) into the tangent equation: \[ \frac{\sqrt{2}}{3}\left(\frac{9}{5}\right) - \frac{\beta}{4} = 1 \] \[ \frac{3\sqrt{2}}{5} - 1 = \frac{\beta}{4} \] \[ \beta = 4\left( \frac{3\sqrt{2}-5}{5} \right) = \frac{12\sqrt{2}-20}{5} \]