Ellipse

Given the ellipse,

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

And assuming that $a > b$ , then the end points of the latus rectum for this ellipse are at, click here for illustration

$(a e, b^2 / a )$ and $( ae, -b^2 / a)$

Back to the question, the given ellipse equation can be rewritten as,

$\dfrac{x^2}{a^2} + \dfrac{y^2}{(ab)^2} = 1$

Assuming $b < 1$ , the end points of its latus rectum are at

$(a e, a b^2 )$ and $(a e , - a b^2 )$

where $e$ is the eccentricity, and for this ellipse, it is given by $e = \sqrt{1 - (\dfrac{ab}{a})^2} = \sqrt{1 - b^2}$

Let $x = ae$ and $y = a b^2$ , then $x^2 + a y = a^2 (1 - b^2) + a^2 b^2 = a^2$ . Hence $m = 2$ . Also, for the other end $(a e , - a b^2)$ , let

$x = a e$ and $y = - a b^2$ , then $x^2 - a y = a^2 (1 - b^2) + a^2 b^2 = a^2$ . Hence $n = 2$ . Therefore, $m - n = 2 - 2 = 0$ .

Note that if $b > 1$ , then the latus rectum endpoints become $( \pm a / b , \pm a b e )$ where $e = \sqrt{ 1 - \dfrac{1}{b^2} }$

In this case, letting $x = a / b$ and $y = a b e$ , then the equation becomes, $y^2 + a b x = (ab)^2 (1 - \dfrac{1}{b^2} ) + a^2 = (ab)^2$ .