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P is a point on the ellipse S=0\, and Q is its corresponding point on the auxiliary circle. Prove that the locus of the point of intersection of the normals at P and Q is the circle given by x^{2}+y^{2}=(a+b)^{2}\,

Equation to the normal at P(\theta )\, on the ellipse is

{\frac  {ax}{\cos \theta }}-{\frac  {by}{\sin \theta }}=a^{2}-b^{2}\,

P1 is the corresponding point on x^{2}+y^{2}=a^{2},P_{1}=1(a\cos \theta ,a\sin \theta )\,

Therefore,equation to the normal to auxiliary circle is y=a\tan \theta \,

From the above,\tan \theta ={\frac  {y}{x}},\sin \theta ={\frac  {x}{{\sqrt  {x^{2}+y^{2}}}}},\cos \theta ={\frac  {y}{{\sqrt  {x^{2}+y^{2}}}}}\,

Eliminating theta from the equation of normal,we get

{\frac  {ax{\sqrt  {x^{2}+y^{2}}}}{x}}-{\frac  {by{\sqrt  {x^{2}+y^{2}}}}{y}}=a^{2}-b^{2}\,

(a-b){\sqrt  {x^{2}+y^{2}}}=a^{2}-b^{2},{\sqrt  {x^{2}+y^{2}}}=a+b\,

Therefore,the locus of the first two equations is x^{2}+y^{2}=(a+b)^{2}\,

Main Page:Geometry:The Ellipse