Question

The pair of lines $6x^2+7xy+\lambda y^2=0(\lambda \ne -6)$ forms a right angled triangle with $x+3y+4=0$ then $\lambda =$

• A. $3$
• B. $-3$
• C. $1/3$
• D. $-1/3$

Answer 1

The correct option is A $3$

Given line is $L:x+3y+4=0$

$⟹y=-\frac{1}{3}(x+4)$

Slope of this line is $m=\frac{-1}{3}$

Now, $6x^2+7xy+\lambda y^2=0$ $(\lambda \ne 6)$

$x^2+\frac{7}{6}xy+\frac{\lambda }{6}{y}^2=0$

$⟹(x+ay)(x+by)=0$

$⟹x+ay=0$ and $x+by=0$ are the two equations with

$a+b=\frac{7}{6}$ and $ab=\frac{\lambda }{6}$

Slope of these lines are $m_1=\frac{-1}{a}$ and $m_2=\frac{-1}{b}$

Now, $m_1{m}_2=\frac{1}{ab}=\frac{\lambda }{6}\ne -1$ since $\lambda \ne -6$

Hence the lines $x+ay=0$ and $x+by=0$ are not prependicular.

From these two only one is normal to $L$.

Let $x+ay$ is normal to $L$.

$⟹m_{1}m=-1$

$⟹\frac{1}{3a}=-1$

$⟹a=\frac{-1}{3}$

Now, $a+b=\frac{7}{6}⟹b=\frac{7}{6}-\frac{-1}{3}$

$⟹b=\frac{3}{2}$

Now, $ab=\frac{\lambda }{6}$

$⟹\lambda =6ab=6\times \frac{-1}{3}\times \frac{3}{2}$

$⟹\lambda =-3$

Answer-(B)