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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of chord of contact of tangents drawn from a point $(x_1,y_1)$ to the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$.

$\begin{array}{1 1}(a)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize =1\\(b)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize =1\\(c)\;\large\frac{xx_1^2}{a^2}+\frac{yy_1^2}{b^2}\normalsize =1\\(d)\;\text{None of these}\end{array}$

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Let PQ and PR be the tangents drawn from a point $P(x_1,y_1)$ to the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
Such that $Q(x',y')$ and $R(x",y")$ are the points of contacts of these tangents the chord QR is called chord of contact of the ellipse.
Equation of tangents at Q(x',y') and R(x",y") are
And $\large\frac{xx"}{a^2}+\frac{yy"}{b^2}$$=1$
These tangents passes through $(x_1,y_1)$
And $\large\frac{x_1x"}{a^2}+\frac{y_1y"}{b^2}$$=1$
$(x',y')$ & $(x",y")$ lie on $\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}$$=1$
Hence equation of QR is
$\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize =1$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v

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