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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the locus of mid-points of the portions of the tangents to the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ intercepted between the axis?

$\begin{array}{1 1}(a)\;a^2y^2+b^2x^2=4x^2y^2\\(b)\;a^2+b^2=4x^2y^2\\(c)\;a^2y^2+b^2x^2=2x^2y^2\\(d)\;\text{None of these}\end{array}$

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Let $P(x_1,y_1)$ be any point on the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$-----(1)
Equation of tangent at $(x_1,y_1)$ is
$\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}$$=1$
This meet the co-ordinate axis at $Q(\large\frac{a^2}{x_1},$$0)$ and $R(0,\large\frac{b^2}{y_1})$
Let $M(h,k)$ be the mid-point of QR then
$h=\large\frac{\Large\frac{a^2}{x_1}+0}{2},$$k=\large\frac{0+\Large\frac{b^2}{y_1}}{2}$
$x_1=\large\frac{a^2}{2h}$,$y_1=\large\frac{b^2}{2k}$
Since $(x_1,y_1)$ lie on equ(1) hence
$\large\frac{(\Large\frac{a^2}{2h})^2}{a^2}+\frac{(\Large\frac{b^2}{2k})^2}{b^2}$$=1$
$\large\frac{a^2}{4h^2}+\frac{b^2}{4k^2}$$=1$
$a^2k^2+b^2h^2=4h^2k^2$
Locus $M(h,k)$ is $a^2y^2+b^2x^2=4x^2y^2$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1
 

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