question_answer

Tangents are drawn to ellipse \[{{x}^{2}}+2{{y}^{2}}=2.\] The locus of mid-point of intercept made by tangents between the axes is -

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 \[\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1\]