Question

Consider the ellipse x2a2+y2b2=1, having it's eccentricity equal to



e. P is any variable point on it and p1p2 are the foot of perpendiculars drawn from p to the x and y-axis respectively. The line p1p2p1p2 will always be a normal to an ellipse whose eccentricity is equal to

Answer 1

$e {}^{2} = \frac{ \frac{1}{ {a}^{2} } - \frac{1}{ {b}^{2} } }{ \frac{1}{ {a}^{2} } }$

${e}^{2} = \frac{ {b}^{2} - {a}^{2} }{ {b}^{2} }$

$e = \sqrt{ \frac{ {b}^{2} - {a}^{2} }{ {b}^{2} } } \: \: if \: \frac{1}{a} > \frac{1}{b}$

ie.b>a.