A) \[{{(4{{x}^{2}}+9{{y}^{2}})}^{2}}=4{{x}^{2}}-9{{y}^{2}}\]
B) \[{{(4{{x}^{2}}-9{{y}^{2}})}^{2}}=4{{x}^{2}}+9{{y}^{2}}\]
C) \[{{(2{{x}^{2}}+3{{y}^{2}})}^{2}}=2{{x}^{2}}-3{{y}^{2}}\]
D) \[{{(2{{x}^{2}}-3{{y}^{2}})}^{2}}=2{{x}^{2}}+3{{y}^{2}}\]
Correct Answer: A
Solution :
Given, equation of hyperbola is \[4{{x}^{2}}-9{{y}^{2}}=1\] Equation of tangent of hyperbola at \[\left( \frac{1}{2}\sec \theta ,\,\,\frac{1}{3}\,\tan \theta \right)\] is \[2x\,\,\sec \theta -3y\,\tan \theta =1\] ?.(i) Let \[(h,k)\] be the mid-point of L, R of the ellipse \[4{{x}^{2}}+9{{y}^{2}}=1\] Since, equation of line passes through \[(h,k),\] which is mid-point of chord LR is \[4hx+9ky=4{{h}^{2}}+9{{y}^{2}}\] \[[\because \,\,T={{S}_{1}}]\] ?.(ii) Now, tangent of hyperbola is intersect the ellipse \[\therefore \] Eqs (i) and (ii) coincide \[\because \] \[\frac{2\,\sec \,\theta }{4h}=\frac{-3\,\tan \theta }{9k}=\frac{1}{4{{h}^{2}}+9{{k}^{2}}}\] \[\Rightarrow \] \[\sec \,\,\theta =\frac{2h}{4{{h}^{2}}+9{{k}^{2}}}\] and \[\tan \theta =\frac{-3k}{4{{h}^{2}}+9{{k}^{2}}}\] On squaring and subtracting, we get \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =\frac{4{{h}^{2}}-9{{k}^{2}}}{{{(4{{h}^{2}}+9{{k}^{2}})}^{2}}}\] \[\Rightarrow \] \[{{(4{{h}^{2}}+9{{k}^{2}})}^{2}}=4{{h}^{2}}-9{{k}^{2}}\] On squaring and subtracting, we get \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =\frac{4{{h}^{2}}-9{{k}^{2}}}{{{(4{{h}^{2}}+9{{k}^{2}})}^{2}}}\] \[\Rightarrow \] \[{{(4{{h}^{2}}+9{{k}^{2}})}^{2}}=4{{h}^{2}}-9{{k}^{2}}\] Hence locus of mid-point of LR is \[{{(4{{x}^{2}}+9{{y}^{2}})}^{2}}=4{{x}^{2}}-9{{y}^{2}}\]
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