Geo3.10

If $ax^2+2hxy+by^2=0\,$ be two sides of a parallelogram and $px+qy=1\,$ is one diagonal,prove that the other diagonal is $y(bp-hq)=x(aq-hp)\,$

Let $P(x_1,y_1),Q(x_2,y_2)\,$ be the points where the diagonal $px+qy=1\,$ meets the pair of lines.

OR and PQ bisect each other at $M(\alpha,\beta)\,$

$\alpha=\frac{x_1+x_2}{2},\beta=\frac{y_1+y_2}{2}\,$

Eliminating y from $ax^2+2hxy+by^2=0,px+qy=1\,$

$ax^2+2hx(\frac{1-px}{q})+b(\frac{1-px}{q})^2=0\,$

$x^2(aq^2-2hpq+bp^2)+2x(hq-bp)+b=0\,$

The roots of this quadratic equation are $x_1,x_2\,$

where $x_1+x_2=-\frac{2(hq-bp)}{aq^2-2hpq-np^2},\alpha=\frac{bp-hq}{aq^2-2hpq+bp^2}\,$

Similarly by eliminating x from the given equations , $y_1+y_2=-\frac{2(hp-aq)}{aq^2-2hpq-np^2},\beta=\frac{aq-hp}{aq^2-2hpq+bp^2}\,$

Now the equation to the line joining origin and M is $\alpha y=\beta x\,$

Hence $y(bp-hq)=x(aq-hp)\,$