# Geo5.2.43

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