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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
A tangent to the hyperbola \(\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1\) meets \(x-\) axis at \(P\) and \(y-\) axis at \(Q\). Lines \(PR\) and \(QR\) are drawn such that \(OPRQ\) is a rectangle (where \(O\) is the origin). Then \(R\) lies on
- A \(\frac{4}{{{x^2}}} + \frac{2}{{{y^2}}} = 1\)
- B \(\frac{2}{{{x^2}}} - \frac{4}{{{y^2}}} = 1\)
- C \(\frac{2}{{{x^2}}} + \frac{4}{{{y^2}}} = 1\)
- D \(\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1\)
Step-by-step Solution
Detailed explanation
equation of the tangent at the point \('\theta '\) is \(\frac{{x\sec \theta }}{a} - \frac{{y\tan \theta }}{b} = 1\) \( \Rightarrow P = \left( {a\cos \theta ,0} \right)\,\,\,\,Q = \left( {0, - b\cot \theta } \right)\) Let \(R\) be…
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