I tried making a diagram of this (please don't look at the numerical values as the tool I used did not let me use the general a, b values. )
So the 2 latus recta here are: BC and FE
and I think what the question asks here is the area of the polygon ABCDEF. Which, is a combination of two trapeziums by symmetry. So I calculated the area of 1 trapezium and multiplied it by 2 and got the final answer:
$**2e(a^2 + b^2)** $
Puttin x=0 in general equation of a tangent through end points of the latus recta, I got the coordinates of A: (0,a) and D: (0,-a)
As the area of trapezium = $\frac{(h)(p+q)}{2}$
taking the trapezium ABCD, I took h= ae as the latus rectum is at a distance of ae from the y-axis, p=BC= $\frac{2b^2}{a}$ and q=AD= 2a
So, the area of trapezium ABCD= $\frac{1}{2}(ae)\left(\frac{2b^2}{a}+2a \right)$
Simplifying and then multiplying it by 2 to get the total area of ABCDEF, I finally got:
$2e(a^2 + b^2)$
But the answer given in my book is: $\frac{b^4}{a \sqrt{a^2-b^2}}$