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Question

A tangent to $$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the axes at $$A$$ and $$B$$. Then the locus of mid point of $${AB}$$ is 


A
a2x2+b2y2=2
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B
a2x2+b2y2=4
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C
a2x2+b2y2=1
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D
a2x2b2y2=12
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Solution

The correct option is B $$\displaystyle \frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$$
Let equation of tangent to the given ellipse is $$ \displaystyle \dfrac{xx_{1}}{a^{2}} +\dfrac{yy_{1}}{b^{2}} = 1$$
Finding its intersection with X and Y axes
$$A = \left(\dfrac{a^{2}}{x_{1}}, 0\right) ;  B = \left(0, \dfrac{b^{2}}{y_{1}}\right)$$
Mid point is $$ \displaystyle \left(\dfrac{a^{2}}{2x_{1}} , \dfrac{b^{2}}{2y_{1}}\right) = (\alpha ,\beta )$$
$$(x_{1}, y_{1})$$ lies on Ellipse
$$ \displaystyle \Rightarrow \dfrac{a^{2}}{4\alpha ^{2}} + \dfrac{b^{2}}{4\beta ^{2}} = 1$$
$$ \displaystyle \therefore \dfrac{a^{2}}{x^{2}} + \dfrac{b^{2}}{y^{2}} = 4$$ is equation of required locus.

Maths

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