Question

If $m$ is the slope of one of the lines represented by $ax^2 + 2hxy + by^2 = 0,$ then $(h + bm)^2$ = ______

• A. $h^2 +ab$
• B. $h^2-ab$
• C. $(a +b)^2$
• D. $(a -b)^2$

Answer 1

Answer: (B)

$h^2-ab$

Answer 2

Given that, $a x^{2}+2 h x y +by^{2}=0$...(i)
Which is homogeneous equation representing pair of straight line each of which passing through the origin. Given one slope of line $=m$.
Let another slope of line $=m_{1}$
Then, the lines are $y=m x$ and $y=m_{1} x$
Now, $(m x-y)\left(m_{1} x-y\right)$
$\Rightarrow m m_{1} x^{2}-m_{1} x y-m x y +y^{2}$
$\Rightarrow m m_{1} \cdot x^{2}-\left(m +m_{1}\right) y \cdot x +y^{2}$...(ii)
On comparing Eqs. (i) and (ii),
$m+m_{1}=-\frac{2 h}{b}$...(iii)
$m m_{1}=\frac{a}{b}$...(iv)
From Eqs. (iii) and (iv),
$m_{1}=\left(-\frac{2 h}{b}-m\right)$
$\Rightarrow m\left(\frac{-2 h}{b}-m\right)=\frac{a}{b}$
$\Rightarrow -\frac{m}{b}(2h +m b)=\frac{a}{b}$
$\Rightarrow -2 m h-m^{2} b=a$
$\Rightarrow -2 m h b-m^{2} b^{2}=a b$
$\Rightarrow h^{2}+2mhb +m^{2} b^{2}=-ab +h^{2}$
$\Rightarrow (h +m b)^{2}=h^{2}-a b$