# Geo5.2.34

Show that the poles of the tangents to the circle $x^{2}+y^{2}=a^{2}+b^{2}\,$ w.r.t the ellipse ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1\,$ lies on ${\frac {x^{2}}{a^{4}}}+{\frac {y^{2}}{b^{4}}}={\frac {1}{a^{2}+b^{2}}}\,$


Any point on the given circle is of the form $P((a^{2}+b^{2})\cos \alpha ,(a^{2}+b^{2})\sin \alpha )\,$

The tangent to the given circle at $P(\alpha )\,$ is

$x\cos \alpha +y\sin \alpha ={\sqrt {a^{2}+b^{2}}}\,$

Let $Q(x_{1},y_{1})\,$ be the pole of this tangent w.r.t teh given ellipse is ${\frac {xx_{1}}{a^{2}}}+{\frac {yy_{1}}{b^{2}}}=1\,$

These two equations represent the same line,hence comparing the coefficients,

${\frac {(a^{2})\cos \alpha }{x_{1}}}={\frac {(b^{2})\sin \alpha }{y_{1}}}={\frac {{\sqrt {a^{2}+b^{2}}}}{1}}\,$

$\cos \alpha ={\frac {x_{1}({\sqrt {a^{2}+b^{2}}})}{a^{2}}},\sin \alpha ={\frac {y_{1}({\sqrt {a^{2}+b^{2}}})}{b^{2}}}\,$

$1=\sin ^{2}\alpha +\cos ^{2}\alpha ={\frac {x_{1}^{{2}}(a^{2}+b^{2})}{a^{4}}}+{\frac {y_{1}^{{2}}(a^{2}+b^{2})}{b^{4}}}\,$
${\frac {x_{1}^{{2}}}{a^{4}}}+{\frac {y_{1}^{{2}}}{b^{4}}}={\frac {1}{a^{2}+b^{2}}}\,$
${\frac {x^{2}}{a^{4}}}+{\frac {y^{2}}{b^{4}}}={\frac {1}{a^{2}+b^{2}}}\,$