If the normal at an end of a latus rectum of an ellipse passes through an extremity of the…

Question

If the normal at an end of a latus rectum of an
ellipse passes through an extremity of the
minor axis, then the eccentricity e of the ellipse
satisfies :

Accepted Answer

Equation of normal at $$\left( {ae,{{{b^2}} \over a}} \right)$$

$${{{a^2}x} \over {ae}} - {{{b^2}y} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$

It passes through (0,--b)

$$ \therefore $$ $$0 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$

$$ \Rightarrow $$ $a$b = ${a^2} - {b^2}$

$$ \Rightarrow $$ $a$b = ${a^2}{e^2}$ [as b2 = $${a^2}\left( {1 - {e^2}} \right)$$]

$$ \Rightarrow $$ $a$2b2 = ${a^4}{e^4}$

$$ \Rightarrow $$ $${{{{b^2}} \over {{a^2}}}}$$ = e4

$$ \Rightarrow $$ ${1 - {e^2}}$ = e4

$$ \Rightarrow $$ e4 + e2 -- 1 = 0

Solutions · 1