Geo5.2.39

Prove that the pair of tangents drawn to $9x^2+16y^2=144\,$ are perpendicular to eachother.

Given $S\equiv 9x^2+16y^2=144\,$ and Let the point be $(x_1,y_1)\,$

$S_1 \equiv 9xx_1+16yy_1-144=0\,$

$S_{11} \equiv 9x_1^{2}+16y_1^{2}-144\,$

Therefore,equation to the pair of tangents is $S_1^{2}=SS_{11}\,$

$(9xx_1+16yy_1-144)^2=(9x^2+16y^2-144)(9x_1^{2}+16y_1^{2}-144)\,$

Separating the coefficients of $x^2\,$ and $y^2\,$

$81x_1^{2}+256y_1^{2}-81x_1^{2}-256y_1^{2}\,$

From the above,the sum of the coefficients of squares of x and y is zero,which is the condition for the two lines to be perpendicular.

Hence the two tangents drawn on the given ellipse are perpendicular to each other.