Question

The line $lx+my+n=0$ intersects the curve $a{x}^2+2hxy+b{y}^2=1$ at the point $P$ and $Q$. The circle on $PQ$ as diameter passes through the origin. Prove that $n^2(a+b)=l^2+m^2$

Answer 1

REF.Image.

curve :- $a{x}^2+2hxy+b{y}^2=1$

line :- $ln+my+n=0\Rightarrow \frac{ln+my}{-n}=1$

homogenizing given curve with the given line

$a{x}^2+2hxy+b{y}^2=(\frac{ln+my}{-n}{)}^2$

$(a-\frac{l^2}{n^2})x^2+(2h-2\frac{l}{n}\frac{m}{n})xy+(b-\frac{m^2}{n^2})y^2=0$__ (1)

$e{q}^n$ (1) give combine $e{q}^n$

of $OA=OB$

as AB is diameter

$\angle AOB={90}^{\circ }$

Thus $m_1{m}^2=-1$

$\downarrow$ $\downarrow$

slope of OB slop of OA

$\left(\frac{y}{x}{)}^2\right(b-\frac{m^2}{n^2})+\frac{y}{x}(2h-\frac{2l}{n}\frac{m}{n})+(a-\frac{l^2}{n^2})=0$

Let $\frac{y}{x}=m$

$m^2(b-\frac{m^2}{n^2})+m(2n-\frac{2l}{n}\frac{m}{n})+(a-\frac{l^2}{n^2})=0$

This quad has 2 roots $m_1$ & $m_2$

we have $m_1{m}_2=-1$

$\frac{a-l^2/n^2}{b-m^2/n^2}=-1$

$\Rightarrow (a+b)n^2=l^2+m^2$