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. Both (a) and (b)

Answer 1

The correct option is D Both (a) and (b)

$y^2+xy-12x^2=0$

$\Rightarrow y^2+4xy-3xy-12y^2=0$

$\Rightarrow y(y+4x)-3x(y+4x)=0$

$\Rightarrow (y+4x)(y-3x)=0$

So, $y=3x$ or $y=-4x$ are the two straight lines represented by the given equation.

We consider two cases, the first being when $y=3x$ is common to both.

Then, we get $a{x}^2+2hx\left(3x\right)+b(3x{)}^2=0$

i.e. $a{x}^2+6h{x}^2+9b{x}^2=0$

$a=-6h-9b=-3(2h+3b)$

The second case being line $y=-4x$ is common to both, which gives

$a{x}^2+2hx(-4x)+b(-4x{)}^2=0$

i.e. $a{x}^2-8h{x}^2+16b{x}^2=0$

$a=8h-16b=8(h-2b)$