Question

The point P\((-2\sqrt{6}, \sqrt{3})\) lies on the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) having eccentricity \(\frac{\sqrt{5}}{2}\). If the tangent and normal at P to the hyperbola intersect its conjugate axis at the points Q and R respectively, then QR is equal to :

• A. \(4\sqrt{3}\)
• B. \(3\sqrt{6}\)
• C. 6
• D. \(6\sqrt{3}\)

Hint:

Memorize the standard forms for the tangent and normal to a hyperbola (and other conics). Tangent at \((x_1, y_1)\) is \(T=0\). Normal is \(\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2+b^2\). The conjugate axis is the y-axis (\(x=0\)) for a standard horizontal hyperbola.

Answer 1

The Correct Option is D

Step 1: Find the Parameters of the Hyperbola
Given eccentricity \(e = \frac{\sqrt{5}}{2}\). We know \(e^2 = 1 + \frac{b^2}{a^2}\). \[ \left(\frac{\sqrt{5}}{2}\right)^2 = 1 + \frac{b^2}{a^2} \implies \frac{5}{4} = 1 + \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{1}{4} \implies a^2 = 4b^2 \] The point P\((-2\sqrt{6}, \sqrt{3})\) lies on the hyperbola, so it satisfies the equation: \[ \frac{(-2\sqrt{6})^2}{a^2} - \frac{(\sqrt{3})^2}{b^2} = 1 \implies \frac{24}{a^2} - \frac{3}{b^2} = 1 \] Substitute \(a^2 = 4b^2\): \[ \frac{24}{4b^2} - \frac{3}{b^2} = 1 \implies \frac{6}{b^2} - \frac{3}{b^2} = 1 \implies \frac{3}{b^2} = 1 \implies b^2 = 3 \] Then \(a^2 = 4(3) = 12\). The equation of the hyperbola is \(\frac{x^2}{12} - \frac{y^2}{3} = 1\).
Step 2: Find the Equation of the Tangent and Point Q
The equation of the tangent at P\((x_1, y_1)\) is \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\). At P\((-2\sqrt{6}, \sqrt{3})\), the tangent is: \[ \frac{x(-2\sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1 \implies -\frac{x\sqrt{6}}{6} - \frac{y\sqrt{3}}{3} = 1 \] The tangent intersects the conjugate axis (\(x=0\)) at point Q. Set \(x=0\): \(-\frac{y_Q\sqrt{3}}{3} = 1 \implies y_Q = -\frac{3}{\sqrt{3}} = -\sqrt{3}\). So, Q = (0, \(-\sqrt{3}\)).
Step 3: Find the Equation of the Normal and Point R
The equation of the normal at P\((x_1, y_1)\) is \(\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2+b^2\). \[ \frac{12x}{-2\sqrt{6}} + \frac{3y}{\sqrt{3}} = 12+3=15 \] \[ -\frac{6x}{\sqrt{6}} + \frac{3y\sqrt{3}}{3} = 15 \implies -x\sqrt{6} + y\sqrt{3} = 15 \] The normal intersects the conjugate axis (\(x=0\)) at point R. Set \(x=0\): \(y_R\sqrt{3} = 15 \implies y_R = \frac{15}{\sqrt{3}} = 5\sqrt{3}\). So, R = (0, \(5\sqrt{3}\)).
Step 4: Calculate the Distance QR
Q = (0, \(-\sqrt{3}\)) and R = (0, \(5\sqrt{3}\)). The distance QR is the difference in their y-coordinates. \[ QR = |5\sqrt{3} - (-\sqrt{3})| = |6\sqrt{3}| = 6\sqrt{3} \] Final Answer: The length of QR is \(6\sqrt{3}\).