Question

If \(C\) is the centre of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] and the tangent drawn at any point \(P\) on the hyperbola meets the lines \[ bx-ay=0 \] and \[ bx+ay=0 \] at \(Q\) and \(R\) respectively, then \[ CQ\cdot CR= \]

• A. \(a^2-b^2\)
• B. \(a^2+b^2\)
• C. \[ \frac1{a^2}+\frac1{b^2} \]
• D. \[ \frac1{a^2}-\frac1{b^2} \]

Hint:

For hyperbola problems involving tangents, use the parametric point \[ (a\sec\theta,\; b\tan\theta) \] and the tangent \[ \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1. \] The identities \[ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1 \] often simplify the final expression dramatically.

Answer 1

The Correct Option is B

Concept: For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the tangent at the point \[ (a\sec\theta,\; b\tan\theta) \] is \[ \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1. \] We find the points where this tangent meets the lines \[ bx-ay=0 \] and \[ bx+ay=0, \] and then compute their distances from the centre \(C(0,0)\).

Step 1: Write the tangent at a parametric point of the hyperbola. At \[ P(a\sec\theta,b\tan\theta), \] the tangent is \[ \frac{x\sec\theta}{a} - \frac{y\tan\theta}{b} = 1. \] \[ b\sec\theta\,x-a\tan\theta\,y=ab. \] \[ \cdots (1) \]

Step 2: Find the point \(Q\) on \(bx-ay=0\). From \[ bx-ay=0, \] \[ y=\frac{b}{a}x. \] Substituting into (1), \[ b\sec\theta\,x - a\tan\theta\left(\frac{b}{a}x\right) = ab. \] \[ bx(\sec\theta-\tan\theta)=ab. \] \[ x=\frac{a}{\sec\theta-\tan\theta}. \] Using \[ (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1, \] \[ x=a(\sec\theta+\tan\theta). \] Hence \[ y=b(\sec\theta+\tan\theta). \] Therefore, \[ Q= \Bigl(a(\sec\theta+\tan\theta), \, b(\sec\theta+\tan\theta)\Bigr). \]

Step 3: Find \(CQ\). Since \(C=(0,0)\), \[ CQ^2 = (\sec\theta+\tan\theta)^2(a^2+b^2). \] Hence, \[ CQ = (\sec\theta+\tan\theta)\sqrt{a^2+b^2}. \]

Step 4: Find the point \(R\) on \(bx+ay=0\). From \[ bx+ay=0, \] \[ y=-\frac{b}{a}x. \] Substituting into (1), \[ b\sec\theta\,x + b\tan\theta\,x = ab. \] \[ bx(\sec\theta+\tan\theta)=ab. \] \[ x=\frac{a}{\sec\theta+\tan\theta}. \] Using \[ \frac1{\sec\theta+\tan\theta} = \sec\theta-\tan\theta, \] \[ x=a(\sec\theta-\tan\theta). \] \[ y=-b(\sec\theta-\tan\theta). \] Thus \[ R= \Bigl(a(\sec\theta-\tan\theta), \, -b(\sec\theta-\tan\theta)\Bigr). \]

Step 5: Find \(CR\). \[ CR^2 = (\sec\theta-\tan\theta)^2(a^2+b^2). \] Therefore, \[ CR = (\sec\theta-\tan\theta)\sqrt{a^2+b^2}. \]

Step 6: Compute \(CQ\cdot CR\). \[ CQ\cdot CR = (\sec\theta+\tan\theta) (\sec\theta-\tan\theta) (a^2+b^2). \] Using \[ (\sec\theta+\tan\theta) (\sec\theta-\tan\theta) = 1, \] we get \[ CQ\cdot CR = a^2+b^2. \]

Step 7: Write the final answer. \[ \boxed{a^2+b^2} \]