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 :
[c] Ellipse \[\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{1}=1\] Let the equation of tangent \[\frac{x}{\sqrt{2}}cos\theta +\frac{y}{1}sin\theta =1\] If (h, k) be the midpoint of the portion of the tangent intercepted between the axes then \[2h=\frac{\sqrt{2}}{\cos \theta },2k=\frac{1}{\sin \theta }\] \[\Rightarrow cos\theta =\frac{1}{\sqrt{2}h}\] and \[\sin \theta =\frac{1}{2k}\] squaring and adding, we get \[\Rightarrow \frac{1}{2{{h}^{2}}}+\frac{1}{4{{k}^{2}}}=1\] \[\therefore \] locus \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]You need to login to perform this action.
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