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Show that the poles of normal chords of the ellipse{\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\, lie on the curve {\frac  {a^{6}}{x^{2}}}+{\frac  {b^{6}}{y^{2}}}=(a^{2}-b^{2})^{2}\,

The equation of normal to the ellipse S=0 at any point theta is

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

Let the pole of l=0 w.r.t S=0 be P(x_{1},y_{1})\,

The polar of P w.r.t S=0 is

{\frac  {xx_{1}}{a^{2}}}+{\frac  {yy_{1}}{b^{2}}}-1=0\,

As these two equations represent the same line i.e polar of P,

Comparing the coefficients, we have

{\frac  {x_{\cos }\theta }{a^{3}}}={\frac  {-y_{1}\sin \theta }{b^{3}}}={\frac  {1}{a^{2}-b^{2}}}\,

\sin \theta ={\frac  {-b^{3}}{y_{1}(a^{2}-b^{2})}},\cos \theta ={\frac  {a^{3}}{x_{1}(a^{2}-b^{2})}}\,

Squaring and adding

1=\sin ^{2}\theta +\cos ^{2}\theta ={\frac  {a^{6}}{x_{1}^{{2}}(a^{2}-b^{2})^{2}}}+{\frac  {b^{6}}{y_{1}^{{2}}(a^{2}-b^{2})^{2}}}\,

{\frac  {a^{6}}{x_{1}^{{2}}}}+{\frac  {b^{6}}{y_{1}^{{2}}}}=(a^{2}-b^{2})^{2}\,

Therefore the pole lies on the curve {\frac  {a^{6}}{x^{2}}}+{\frac  {b^{6}}{y^{2}}}=(a^{2}-b^{2})^{2}\,

Main Page:Geometry:The Ellipse