Question

If the two pair of lines $2x^2+6xy+y^2=0$ and $4x^2+18xy+b{y}^2=0$ are equally inclined, then $b$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (A)

Given pair of lines are

$2x^2+6xy+y^2=0$

$4x^2+18xy+b{y}^2=0$

We know that, Equation of angle bisector of $a{x}^2+2hxy+b{y}^2=0$ is $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

Now, lets take, $2x^2+6xy+y^2=0$

Here, $a=2,b=1$

and $2h=6\Rightarrow h=3$

Equation of angle bisector of equation is $\frac{x^2-y^2}{2-1}=\frac{xy}{3}$

$\Rightarrow \frac{x^2-y^2}{xy}=\frac{1}{3}$---(1)

Lets take $4x^2+18xy+b{y}^2=0$

Here, $a=4,b=b$ and $2h=18\Rightarrow h=9$

Equation of angle bisector of equation is $\frac{x^2-y^2}{4-b}=\frac{xy}{9}$

$\Rightarrow \frac{x^2-y^2}{xy}=\frac{4-b}{3}$---(2)

It is given that, $2x^2+6xy+y^2=0$ and $4x^2+18xy+b{y}^2=0$ are equally inclined.

i.e., from (1) and (2

$\frac{x^2-y^2}{xy}=\frac{1}{3}=\frac{4-b}{9}$

$\Rightarrow \frac{4-b}{9}=\frac{1}{3}$

$\Rightarrow 4-b=3$

$\Rightarrow b=1$

Hence, Option A is correct.