Originally Posted by

**jiboom** i need to show

$\displaystyle

-1\leq \frac{-2b^2}{a^2+b^2}\leq0

$

now i have shown

$\displaystyle e^2=1+\frac{b^2}{a^2}\leq2$

which gives me $\displaystyle b^2\leq a^2$

clearly the expression is $\displaystyle \leq 0 $ so i need to show

$\displaystyle -1\leq \frac{-2b^2}{a^2+b^2}$

this is simple enough by considering

$\displaystyle

\frac{-2b^2}{a^2+b^2}+1 [*]

$

$\displaystyle

= \frac{a^2-b^2}{a^2+b^2}

$

as $\displaystyle b^2\leq a^2$

$\displaystyle a^2-b^2 > 0$

so [*]>0

What i would like to do is not use the result and start from scratch to show

$\displaystyle

-1\leq \frac{-2b^2}{a^2+b^2} $

using only

$\displaystyle

e^2=1+\frac{b^2}{a^2}\leq2

$

i cant seem to get the result, alwayus end up with an inequality round the wrong way so cant deduce the result. Maybe it cant be done this way which is why they gave the value -1 or im making a mistake: