A) \[\frac{1}{{{x}^{2}}}+\frac{1}{2{{y}^{2}}}=1\]
B) \[\frac{1}{4{{x}^{2}}}+\frac{1}{2{{y}^{2}}}=1\]
C) \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]
D) \[\frac{1}{2{{x}^{2}}}+\frac{1}{{{y}^{2}}}=1\]
Correct Answer: C
Solution :
Let the point of contact be \[R\equiv (\sqrt{2}\cos \theta ,\,\sin \theta )\] Equation of tangent AB is \[\frac{x}{\sqrt{2}}\cos \theta +y\sin \theta =1\] \[\Rightarrow \]\[A\equiv (\sqrt{2}\sec \theta ,\,0);\,B\equiv (0,\,\text{cosec }\theta )\] Let the middle point Q of AB be \[(h,\,k)\] \[\Rightarrow \] \[h=\frac{\sec \theta }{\sqrt{2}},\,k=\frac{\text{cosec }\theta }{2}\Rightarrow \cos \theta =\frac{1}{h\sqrt{2}},\,\sin \theta =\frac{1}{2k}\] \[\Rightarrow \] \[\frac{1}{2{{h}^{2}}}+\frac{1}{4{{k}^{2}}}=1\], \Required locus is \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]. Trick : The locus of mid-points of the portion of tangents to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] intercepted between axes is \[{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}=4{{x}^{2}}{{y}^{2}}\] i.e., \[\frac{{{a}^{2}}}{4{{x}^{2}}}+\frac{{{b}^{2}}}{4{{y}^{2}}}=1\] or \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\].You need to login to perform this action.
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