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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
A normal to the hyperbola, \(4x^2 - 9y^2\, = 36\) meets the co-ordinate axes \(x\) and \(y\) at \(A\) and \(B\), respectively . If the parallelogram \(OABP\) ( \(O\) being the origin) is formed, then the locus of \(P\) is
- A \(4x^2 -9y^2\, = 121\)
- B \(4x^2 +9y^2\,= 121\)
- C \(9x^2 -4y^2\, = 169\)
- D \(9x^2 +4y^2\, = 169\)
Answer & Solution
Correct Answer
(C) \(9x^2 -4y^2\, = 169\)
Step-by-step Solution
Detailed explanation
Given, \(4{x^2} - 9{y^2} = 36\) ater differentiating w.r.t.\(x\), we get \(4.2x - 9.2y.\frac{{dy}}{{dx}} = 0\) \( \Rightarrow \) slope of tangent \( = \frac{{dy}}{{dx}} = \frac{{4x}}{{9y}}\) so, slope of normal \( = \frac{{ - 9y}}{{4x}}\) Now, equation of normal at point…
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