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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}}}\,

Squaring and adding,we get

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}}}\,

Hence the locus of the poles of the tangetns of the circle w.r.t the ellipse is

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

Main Page:Geometry:The Ellipse