For the ellipse frac{x^2}{9} + frac{y^2}{4} = | vedclass

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(C) Let the point of intersection be $R(x_1, y_1)$. Then $PQ$ is the chord of contact of $R$ with respect to the ellipse,and its equation is $\frac{xx

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Ellipse

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(C) Let the point of intersection be $R(x_1, y_1)$. Then $PQ$ is the chord of contact of $R$ with respect to the ellipse,and its equation is $\frac{xx_1}{9} + \frac{yy_1}{4} = 1$.The joint equation of the lines joining $P$ and $Q$ to the center $O(0, 0)$ is obtained by homogenizing the ellipse equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$ using the chord of contact equation: $\frac{x^2}{9} + \frac{y^2}{4} = (\frac{xx_1}{9} + \frac{yy_1}{4})^2$.Since $OP \perp OQ$,the sum of the coefficients of $x^2$ and $y^2$ must be zero.Expanding the equation: $\frac{x^2}{9} + \frac{y^2}{4} = \frac{x^2x_1^2}{81} + \frac{y^2y_1^2}{16} + \frac{2xyx_1y_1}{36}$.Rearranging: $x^2(\frac{x_1^2}{81} - \frac{1}{9}) + y^2(\frac{y_1^2}{16} - \frac{1}{4}) + \frac{2xyx_1y_1}{36} = 0$.Setting the sum of coefficients of $x^2$ and $y^2$ to zero: $(\frac{x_1^2}{81} - \frac{1}{9}) + (\frac{y_1^2}{16} - \frac{1}{4}) = 0$.Thus,the locus of $(x_1, y_1)$ is $\frac{x^2}{81} + \frac{y^2}{16} = \frac{1}{9} + \frac{1}{4} = \frac{13}{36}$,which is an ellipse.

For the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,a chord $PQ$ subtends a right angle at the center. What is the locus of the point of intersection of the tangents at $P$ and $Q$?

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