Question

If the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents a pair of straight lines, then find the square of the distance of their point of intersection from the origin

Answer 1

The general solution is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ ....$\left(1\right)$

Let $(\alpha ,\beta )$ be the point of intersection we consider line paralleled transformation.

Let $x=x^{\prime }+\alpha ,y=y^{\prime }+\beta$

From $\left(1\right)$ we have

$a{(x^{\prime }+\alpha )}^2+2h(x^{\prime }+\alpha )(y^{\prime }+\beta )+b{(y^{\prime }+\beta )}^2+2g(x^{\prime }+\alpha )+2f(y^{\prime }+\beta )+c=0$

$\Rightarrow a{x^{\prime }}^2+2h{x}^{\prime }{y}^{\prime }+b{y^{\prime }}^2+a{\alpha }^2+2h\alpha \beta +b{\beta }^2+2g\alpha +2f\beta +2x^{\prime }(a\alpha +h\beta +g)+2y^{\prime }+2y^{\prime }(h\alpha +b\beta +f)=0$

$\Rightarrow a{x^{\prime }}^2+2h{x}^{\prime }{y}^{\prime }+b{y^{\prime }}^2+2x^{\prime }(a\alpha +h\beta +g)+2g^{\prime }+2y^{\prime }(h\alpha +b\beta +f)$

which is of the form $a{x^{\prime }}^2+2h{x}^{\prime }{y}^{\prime }+b{y^{\prime }}^2=0$

This cannot be possible unless

$a\alpha +h\beta +g=0$ and $h\alpha +b\beta +f=0$

Solving we get

$\frac{a}{hf-bg}=\frac{\beta }{hg-af}=\frac{1}{ab-h^2}$

$\therefore a=\frac{hf-bg}{ab-h^2}$ and $\beta =\frac{hg-af}{ab-h^2}$