WBJEE · Maths · Hyperbola
Let \(\mathrm{P}(3 \sec \theta, 2 \tan \theta) \mathrm{Q}(3 \sec \phi, 2 \tan \phi)\) be two points on \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1\) such that \(\theta+\phi=\frac{\pi}{2} .0 < \theta, \phi < \frac{\pi}{2}\). Then the ordinate of the point of intersection of the normals at \(P\) and \(Q\) is
- A \(\frac{13}{2}\)
- B \(-\frac{13}{2}\)
- C \(\frac{5}{2}\)
- D \(-\frac{5}{2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{13}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2, P(3 \sec \theta, 2 \tan \theta) \\ \therefore & \frac{9 x}{3 \sec \theta}+\frac{4 y}{2 \tan \theta}=9+4=13 \\ \Rightarrow & 3 x \cos \theta+2 y \cot \theta=13 \end{aligned}\) Similarly,…
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