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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}}}\,

Squaring and adding these two

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


{\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}}}\,

Main Page:Geometry:The Ellipse