AP EAMCET · Maths · Hyperbola
Let \(C\) be the centre of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and \(P\) be a point on it. If the tangent at \(P\) to the hyperbola meets the straight lines \(b x-a y=0\) and \(b x+a y=0\) respectively in \(Q\) and \(R\), then \(C Q . C R=\)
- A \(a^2-b^2\)
- B \(a^2+b^2\)
- C \(\frac{1}{a^2}-\frac{1}{b^2}\)
- D \(\frac{1}{a^2}+\frac{1}{b^2}\)
Answer & Solution
Correct Answer
(B) \(a^2+b^2\)
Step-by-step Solution
Detailed explanation
Let \(P=(a \sec\theta, b \tan\theta)\) be a point on the hyperbola. The tangent at \(P\) is \(\frac{x \sec\theta}{a} - \frac{y \tan\theta}{b} = 1\). The asymptotes are \(y=\frac{b}{a}x\) and \(y=-\frac{b}{a}x\). For \(Q\) (intersection with \(y=\frac{b}{a}x\)):…
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