A) \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]
B) \[\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{4}=1\]
C) \[\frac{1}{4{{x}^{2}}}+\frac{1}{2{{y}^{2}}}=1\]
D) \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{2}=1\]
Correct Answer: A
Solution :
Given equation of ellipse is \[\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{1}=1\] Here, \[a=\sqrt{2}\]and b = 1 \[\therefore \] Equation of tangent to the given ellipse is \[\frac{x}{a\sec \theta }+\frac{y}{b\,\cos ec\theta }=1\] \[\Rightarrow \]\[\frac{x}{\sqrt{2}\sec \theta }+\frac{y}{\cos ec\theta }=1\] Let the midpoint of the tangent intercepted between the axes be (h, k). \[\therefore \]\[h=\frac{\sqrt{2}\sec \theta }{2}\Rightarrow \cos \theta =\frac{1}{\sqrt{2}h}\] and\[k=\frac{co\sec \theta }{2}\Rightarrow \sin \theta =\frac{1}{2k}\] Since,\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] \[\Rightarrow \]\[\frac{1}{4{{k}^{2}}}+\frac{1}{2{{h}^{2}}}=1\]\[\Rightarrow \]\[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]
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