A chord PQ of the ellipse frac{x^2}{9} + frac | vedclass

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(C) Let the point of intersection be $R(x_1, y_1)$. The equation of the chord of contact $PQ$ is $\frac{xx_1}{9} + \frac{yy_1}{4} = 1$.To find the com

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an ellipse

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(C) Let the point of intersection be $R(x_1, y_1)$. The equation of the chord of contact $PQ$ is $\frac{xx_1}{9} + \frac{yy_1}{4} = 1$.To find the combined equation of lines $OP$ and $OQ$,we homogenize the ellipse equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$ using the chord of contact:$\frac{x^2}{9} + \frac{y^2}{4} = (\frac{xx_1}{9} + \frac{yy_1}{4})^2$.Since $OP \perp OQ$,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(\frac{1}{9} - \frac{x_1^2}{81}) + (\frac{1}{4} - \frac{y_1^2}{16}) = 0$.Rearranging the terms,we get $\frac{x_1^2}{81} + \frac{y_1^2}{16} = \frac{1}{9} + \frac{1}{4} = \frac{13}{36}$.Thus,the locus of $(x_1, y_1)$ is $\frac{x^2}{81} + \frac{y^2}{16} = \frac{13}{36}$,which represents an ellipse.

$A$ chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends a right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

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