Question:

If the normal to the rectangular hyperbola $x^2 - y^2 = 1$ at the point $P(\tfrac{\pi}{4})$ meets the curve again at $Q(\theta)$, then $\sec^2\theta + \tan\theta =$

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Use the parametric form for rectangular hyperbolas and geometric interpretation of normals.
Updated On: May 19, 2025
  • $43$
  • $57$
  • $3$
  • $1$
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The Correct Option is B

Solution and Explanation

Standard parametric form of rectangular hyperbola $x^2 - y^2 = 1$ is $x = \sec\theta, y = \tan\theta$
Given $P = (\sec(\tfrac{\pi}{4}), \tan(\tfrac{\pi}{4})) = (\sqrt{2}, 1)$
Find equation of normal and second point of intersection $Q(\theta)$
From identity and computation we get $\sec^2\theta + \tan\theta = 57$
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