Question

Two pairs of straight lines have the equations $y^2+xy-12x^2=0$ and $a{x}^2+2hxy+b{y}^2=0$.One line will be common among them, if

• A. $a=-3(2h+3b)$
• B. $a=8(h-2b)$
• C. $a=2(b+h)$
• D. $a=-3(b+h)$

Answer 1

Answer: (A), (B)

$a=-3(2h+3b)$
$a=8(h-2b)$

$y^2+xy-12x^2=0$ ...........$\left(1\right)$
On factorizing, we get $(y+4x)(y-3x)=0$

$\therefore \frac{y}{x}=-4,3$

$a{x}^2+2hxy+b{y}^2=0$
Divide the above equation by $x^2$ we get

$\Rightarrow a+2h\left(\frac{y}{x}\right)+b{\left(\frac{y}{x}\right)}^2$ ...............$\left(2\right)$

The two pairs will have a line common, if $-4$ or $3$ is a root of eqn$\left(2\right)$

Then,substituting for $\frac{y}{x}=-4,3$ in eqn$\left(2\right)$ we get

when $\frac{y}{x}=-4$ ,we have $a-8h+16b=0$

when $\frac{y}{x}=3$ we have $a+6h+9b=0$