A parallelogram is formed by the lines $ax^2+2hxy+by^2=0$ and the lines through $(p,q)$ parallel to them and the equation of the diagonal of the parallelogram which doesn't pass through the origin is $(\lambda x-p)(ap+hq)+(\phi y-q)(hp-bp)=0$ then find the value of $\lambda^3+\phi^3$
I have found the slopes of the lines by comparing the coefficients of the various terms in $(m_1x-y+c_1)(m_2x-y+c_2)=0$ and the given equation for the $2$ lines. Hence I obtain $m_1 m_2=\frac{a}{b}$ and $m_1+m_2=-2h$ which can be solved to get the slopes.
$$m_1=\frac{-h-\sqrt{h^2-4ab}}{2}$$
$$m_2=\frac{-h+\sqrt{h^2-4ab}}{2}$$
After this I have tried to find the intersection of the lines to obtain the opposite points connecting the required diagonal. The points I obtain are quite complex and seem to be incorrect. My procedure also seems quite long.
Is there a shorter method to obtain the answer?
$h^2-ab>0$
Hyperbola passes through $\left( \frac{p}{2}, \frac{q}{2}\right)$ $$4(ax^2+2hxy+by^2)=ap^2+2hpq+bq^2$$
Asymptotes $$ax^2+2hxy+by^2=0$$
Tangent at $(x',y')$
$$4[ax'x+h(y'x+x'y)+by'y]=ap^2+2hpq+bq^2$$
Two diagonals bisect each other at $\left( \frac{p}{2}, \frac{q}{2}\right)$.
The tangent at $\left( \frac{p}{2}, \frac{q}{2}\right)$ is the diagonal not passing through the origin, i.e.
$$2[apx+h(qx+py)+bqy]=ap^2+2hpq+bq^2$$
$$2(ap+hq)x+2(hp+bq)y=p(ap+hq)+q(hp+bq)$$
$$(ap+hq)(2x-p)+(hp+bq)(2y-q)=0$$
$$\lambda=\phi=2$$
$$\fbox{$\lambda^3+\phi^3=16$}$$
See another answer here for your interest.