Question

If the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents a pair of straigth lines,

then the square of the distances of their point of intersection from the origin

is?

• A. $\frac{c(a+b)-a{f}^2-b{g}^2}{ab-h^2}$
• B. $\frac{c(a+b)+f^2+g^2}{ab-h^2}$
• C. $\frac{c(a+b)-f^2-g^2}{ab-h^2}$
• D. $\frac{c(a+b)-f^2-g^2}{(ab-h^2{)}^2}$

Answer 1

Answer: (C)

$\frac{c(a+b)-f^2-g^2}{ab-h^2}$

Pt of intersection of the lines represented by $a{x}^2+2hxy+b{y}^2+2gx+2fg+c=0$

is $(\sqrt{\frac{f^2-ac}{h^2-ab}},\sqrt{\frac{g^2-ca}{h^2-ab}})$

$\therefore$ distance from origin is

$d=\sqrt{\frac{f^2-bc}{h^2-ab}+\frac{g^2-ac}{h^2-ab}}$

$d=\sqrt{\frac{g^2+f^2-c(a+b)}{h^2-ab}}$

$\therefore d^2=\frac{g^2+f^2-c(a+b)}{h^2-ab}$