# Geo5.2.35

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

Any point on the ausiliary circle is of the form $P(a\cos \alpha ,a\sin \alpha )\,$

The tangent to the circle at the point $P(\alpha )\,$ is $x\cos \alpha +y\sin \alpha =a\,$

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

As these two equations represent the same line,comparing the coefficients,we have

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

$\cos \alpha ={\frac {x_{1}}{a}},\sin \alpha ={\frac {ay_{1}}{b^{2}}}\,$

$1={\frac {x_{1}^{{2}}}{a^{2}}}+{\frac {a^{2}y_{1}^{{2}}}{b^{4}}}\,$

Therefore,

${\frac {x_{1}^{{2}}}{a^{4}}}+{\frac {y_{1}^{{2}}}{b^{4}}}={\frac {1}{a^{2}}}\,$

Therefore,the point Q lies on the curve ${\frac {x^{2}}{a^{4}}}+{\frac {y^{2}}{b^{4}}}={\frac {1}{a^{2}}}\,$