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: Understanding the Concept
This problem requires finding the intersection point of the tangent to a hyperbola at a given point and one of its directrices. We need the formulas for the equation of the tangent and the equation of the directrix.
Step 2: Key Formula or Approach
1. Identify the parameters $a$ and $b$ for the hyperbola. 2. Write the equation of the tangent to the hyperbola at the given point $(x_1, y_1)$ using the formula $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. 3. Calculate the eccentricity $e = \sqrt{1+\frac{b^2}{a^2}}$. 4. Write the equation of the directrices, $x=\pm \frac{a}{e}$. 5. Find the intersection point Q by solving the system of equations for the tangent and the appropriate directrix. The problem specifies the fourth quadrant, which helps choose the correct directrix.
Step 3: Detailed Explanation
1. Hyperbola parameters and point P: The hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$. So, $a^2=9 \implies a=3$ and $b^2=16 \implies b=4$. The point is $P(3\sqrt{2}, 4)$. 2. Equation of the tangent at P: Using the formula $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$: \[ \frac{x(3\sqrt{2})}{9} - \frac{y(4)}{16} = 1 \] \[ \frac{\sqrt{2}x}{3} - \frac{y}{4} = 1 \] 3. Equation of the directrix: First, find the eccentricity $e$: \[ e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{16}{9}} = \sqrt{\frac{9+16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \] The equations of the directrices are $x = \pm \frac{a}{e} = \pm \frac{3}{5/3} = \pm \frac{9}{5}$. The point P is on the right branch of the hyperbola ($x=3\sqrt{2}>a=3$), so the relevant tangent will intersect the corresponding directrix, $x=a/e = 9/5$. 4. Find the intersection point Q: We need to find the intersection of the tangent line $\frac{\sqrt{2}x}{3} - \frac{y}{4} = 1$ and the directrix $x = \frac{9}{5}$. Substitute the value of $x$ into the tangent equation: \[ \frac{\sqrt{2}}{3}\left(\frac{9}{5}\right) - \frac{y}{4} = 1 \] \[ \frac{3\sqrt{2}}{5} - \frac{y}{4} = 1 \] \[ \frac{y}{4} = \frac{3\sqrt{2}}{5} - 1 \] \[ \frac{y}{4} = \frac{3\sqrt{2}-5}{5} \] \[ y = \frac{4(3\sqrt{2}-5)}{5} = \frac{12\sqrt{2}-20}{5} \] So, the intersection point is $Q(\alpha, \beta) = \left(\frac{9}{5}, \frac{12\sqrt{2}-20}{5}\right)$. The question asks for the value of $\beta$. \[ \beta = \frac{12\sqrt{2}-20}{5} \] We must verify that Q is in the fourth quadrant. The x-coordinate $\alpha = 9/5$ is positive. For the y-coordinate $\beta$, we compare $12\sqrt{2}$ and $20$. Squaring them gives $(12\sqrt{2})^2 = 144 \times 2 = 288$ and $20^2 = 400$. Since $288<400$, $12\sqrt{2}<20$, which means $12\sqrt{2}-20$ is negative. So $\beta<0$. The point Q is indeed in the fourth quadrant. Step 4: Final Answer
The y-coordinate of the intersection point Q is $\beta = \frac{12\sqrt{2}-20}{5}$.