Question:
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 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 :
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Write the hyperbola as \(\frac{x^2}{9}-\frac{y^2}{4}=1\), use the normal at \((3\sec\theta, 2\tan\theta)\), and note P = A + B.
Updated On: Nov 11, 2026
- \(4x^2+9y^2=121\)
- \(9x^2+4y^2=169\)
- \(4x^2-9y^2=121\)
- \(9x^2-4y^2=169\)
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The Correct Option is D
Solution and Explanation
Hyperbola: \(\frac{x^2}{9}-\frac{y^2}{4}=1\), so \(a=3,b=2,a^2+b^2=13\).
Normal at \((3\sec\theta,2\tan\theta)\): \(\frac{3x}{\sec\theta}+\frac{2y}{\tan\theta}=13\), i.e. \(3x\cos\theta+2y\cot\theta=13\).
Put y = 0: \(A=\left(\frac{13}{3\cos\theta},0\right)\). Put x = 0: \(B=\left(0,\frac{13\tan\theta}{2}\right)\).
OABP is a parallelogram, so P = A + B = \(\left(\frac{13\sec\theta}{3},\frac{13\tan\theta}{2}\right)=(h,k)\).
Then \(\sec\theta=\frac{3h}{13}\), \(\tan\theta=\frac{2k}{13}\). Using \(\sec^2\theta-\tan^2\theta=1\): \(\frac{9h^2}{169}-\frac{4k^2}{169}=1\).
Locus: \(9x^2-4y^2=169\).
Answer: (4) \(9x^2-4y^2=169\)
Normal at \((3\sec\theta,2\tan\theta)\): \(\frac{3x}{\sec\theta}+\frac{2y}{\tan\theta}=13\), i.e. \(3x\cos\theta+2y\cot\theta=13\).
Put y = 0: \(A=\left(\frac{13}{3\cos\theta},0\right)\). Put x = 0: \(B=\left(0,\frac{13\tan\theta}{2}\right)\).
OABP is a parallelogram, so P = A + B = \(\left(\frac{13\sec\theta}{3},\frac{13\tan\theta}{2}\right)=(h,k)\).
Then \(\sec\theta=\frac{3h}{13}\), \(\tan\theta=\frac{2k}{13}\). Using \(\sec^2\theta-\tan^2\theta=1\): \(\frac{9h^2}{169}-\frac{4k^2}{169}=1\).
Locus: \(9x^2-4y^2=169\).
Answer: (4) \(9x^2-4y^2=169\)
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