If $ax^2+2hxy+by^2$ be the two sides of the parallelogram and $px+qy=1$ is one diagonal then prove that the other diagonal is $y(bp-hq)=x(aq-hp)$.
By reading the question I just understood that the given diagonal is not through origin. Second diagonal is through origin and contains the mid point of the given diagonal but could not proceed further
If you solve the diagonal equation simultaneously withvthe pair of sraight lines, eliminating $y$, you get the quadratic in $x$: $$x^2(aq^2-2hpq+bp^2)+x(2hq-2pb)+b=0$$
Considering the sum of roots, the midpoint of the diagonal has $x$ coordinate $$\frac{x_1+x_2}{2}=\frac{pb-hq}{aq^2-2hpq+bp^2}$$
Likewise, the $y$ coordinate can be deduced directly by swapping $p$ for $q$ and $a$ for $b$, resulting in: $$\frac{y_1+y_2}{2}=\frac{aq-hp}{aq^2-2hpq+bp^2}$$
Now you can write down the gradient of the other diagonal which passes though the origin and the result follows immediately.
