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Prove that the sum of the squares of the perpendiculars on any tangent of the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\,(a > b) from two points on the minor axis,each at a distance of {\sqrt  {a^{2}-b^{2}}}\, from the centre is 2a^{2}\,

The equation of the tangents to the given ellipse at P(x_{1},y_{1})\, is

y_{1}=mx_{1}+{\sqrt  {a^{2}m^{2}+b^{2}}}\,

The two given points are (0,{\sqrt  {a^{2}-b^{2}}}),(0,-{\sqrt  {a^{2}-b^{2}}})\,

The perpendicular distances from the two points are

{\frac  {|-{\sqrt  {a^{2}-b^{2}}}+{\sqrt  {a^{2}m^{2}+b^{2}}}|}{{\sqrt  {1+m^{2}}}}},{\frac  {|{\sqrt  {a^{2}-b^{2}}}+{\sqrt  {a^{2}m^{2}+b^{2}}}|}{{\sqrt  {1+m^{2}}}}}\,

The sum of the squares of these two is

{\frac  {2(a^{2}-b^{2}+a^{2}m^{2}+b^{2})}{1+m^{2}}}\,

{\frac  {2a^{2}(1+m^{2})}{1+m^{2}}}\,


Main Page:Geometry:The Ellipse