Geo5.2.19

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If the normal at one end of a latus rectum of the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\, passes through one end of the minor axis,then show that e^{4}+e^{2}=1\,

Let L be the one end of the latus rectum of the given ellipse.Then the coordinates of L are

\left(ae,{\frac  {b^{2}}{a}}\right)\,.

Hence the equation of the normal of L is {\frac  {a^{2}x}{ae}}-{\frac  {ab^{2}y}{b^{2}}}=a^{2}-b^{2}\,

Thi sline passes through the one end (0,-b) of minor axis of the given ellipse.

{\frac  {a(0)}{e}}-a(-b)=a^{2}-b^{2}\,

ab=a^{2}-a^{2}(1-e^{2})\,

ab=a^{2}e^{2}\,

e^{2}={\frac  {b}{a}}\,

Therefore

e^{4}={\frac  {b^{2}}{a^{2}}}={\frac  {a^{2}(1-e^{2})}{a^{2}}}\,

Hencee^{4}+e^{2}=1\,

Main Page:Geometry:The Ellipse