# Geo5.2.38

A chord PQ of an ellipse subtends a right angle at the centre of the ellipse $S=0\,$.Show that the point of intersection of tangents at P and Q lies on the ellipse$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}\,$

Let the tangents at P and Q intersects at $R(x_1,y_1)\,$,then PQ is the chord of tangents drawn from $(x_1,y_1)\,$

Therefore $PQ=\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1\,$

The equation of the pair of lines CP and CQ is obtained by homogenising the equation of S=0 with PQ.

The combined equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=[\frac{xx_1}{a^2}+\frac{yy_1}{b^2}]^2\,$

$x^2[\frac{1}{a^2}-\frac{x_1^{2}}{a^4}]-\frac{2x_1 y_1 xy}{a^2 b^2}+y^2[\frac{1}{b^2}-\frac{y_1^{2}}{b^4}]=0\,$

But angle PCQ is 90.

Therefore,the coefficients of square of x and square of y is equal to zero.

$[\frac{1}{a^2}-\frac{x_1^{2}}{a^4}]+[\frac{1}{b^2}-\frac{y_1^{2}}{b^4}]=0\,$

$\frac{x_1^{2}}{a^4}+\frac{y_1^{2}}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}\,$

Hence the point of intersection of tangents at P and Q lies on the ellipse

$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}\,$

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