Geo5.2.20

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Show that the points of intersection of the perpendicular tangents to an ellipse lies on a circle.

Let the equation of ellipse be {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1,a>b\,

Any tangent to it is y=mx+{\sqrt  {a^{2}m^{2}+b^{2}}}\,

Let the perpendicular tangents intersect at P(x_{1},y_{1})\,

Therefore,P lies on the tangent for some real m.

(y_{1}-mx_{1})^{2}=a^{2}m^{2}+b^{2}\,

In another form,

(x_{1}^{{2}}-a^{2})m^{2}-2x_{1}y_{1}m+(y_{1}^{{2}}-b^{2})=0\, is a quadratic equation in m. Let

its roots be m1,m2.Then,m1,m2 are the slopes of tangents from P to the ellipse.

Therefore,\tan \theta _{1}\tan \theta _{2}=m_{1}m_{2}={\frac  {y_{1}^{{2}}-b^{2}}{x_{1}^{{2}}-a^{2}}}\,

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

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

Therefore,the point of intersection of perpendicular tangents to the ellipse lies on the circle

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


Main Page:Geometry:The Ellipse