Geo5.2.36

From Example Problems
Jump to: navigation, search

Prove that the product of the perpendicular from the foci on any tangent to the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\, is equal to b^{2}\,

Given ellipse is {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\,

Let the equation of any tangent to the ellipse be y=mx+{\sqrt  {a^{2}m^{2}+b^{2}}}\,

SL=Length of the perpendicular from S(ae,0) on the tangent line above

{\frac  {|m(ae)+{\sqrt  {a^{2}m^{2}+b^{2}}}|}{{\sqrt  {m^{2}+1}}}}\,

Length of the perpendicular from S(-ae,0) to the tangent is {\frac  {|m(-ae)+{\sqrt  {a^{2}m^{2}+b^{2}}}|}{{\sqrt  {m^{2}+1}}}}\,

Product of these two perpendiculars is {\frac  {a^{2}m^{2}+b^{2}-a^{2}e^{2}m^{2}}{m^{2}+1}}={\frac  {m^{2}a^{2}(1-e^{2})+b^{2}}{m^{2}+1}}\,

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


Main Page:Geometry:The Ellipse