Question

Show that the lines joning the origin to the points common to $x^2+hxy-y^2+gx+fy=0$ and $fx-gy=\lambda$ are at right angles whatever be the value of $\lambda$.

Answer 1

$x^2+hxy-y^2+9x+fy=0$ ………..$\left(1\right)$

$f\left(x\right)=gy=\lambda$ …………..$\left(2\right)$

$x^2+hxy-y^2+9x=-fy$ ………..$\left(3\right)$

$fx-gy=\lambda$ ……….$\left(4\right)$

From $\left(4\right)$

$\frac{fx-gy}{\lambda }=1$

$x^2+hxy-y^2+9x\left(\frac{fx-gy}{\lambda }\right)+fy\left(\frac{fx-gy}{\lambda }\right)=0$

$\lambda x^2+1hxy-\lambda y^2+fg{x}^2-g^{2}xy+f^{2}xy-fg{y}^2=0$

$x^2(\lambda +fg)+xy(\lambda h+f^2-g^2)+y^2(-\lambda -fg)=0$ ……$\left(5\right)$

This is the equation of line joining the origin and the point common to line $\left(1\right)$ and line $\left(2\right)$.

Comparing $\left(5\right)$ with

$a{x}^2+2hxy+b{y}^2=0$

We getr

$a=\lambda +fg$

$h=\frac{\lambda h+f^2-g^2}{2}$

$b=(-\lambda -fg)$

$\tan \theta =\left|\frac{2\sqrt{h^2+ab}}{a+b}\right|=\left|\frac{2\sqrt{h^2+ab}}{0}\right|$

$theta=ta{n}^{-1}\left(\infty \right)={90}^o$.