Question

Find the locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}\lambda =0$ and $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

• A. $4x^2-y^2=48$
• B. $x^2-4y^2=48$
• C. $3x^2-y^2=48$
• D. $y^2-3x^2=48$

Answer 1

The correct option is C $3x^2-y^2=48$

Let $(h,k)$ be the point of intersection of the given lines.

Then,

$\sqrt{3}h-k-4\sqrt{3}\lambda =0$ and $\sqrt{3}\lambda h+\lambda k-4\sqrt{3}=0$

$\sqrt{3}h-k=4\sqrt{3}\lambda$ and $\lambda (\sqrt{3}h+k)=4\sqrt{3}$

multiplying both the equations

$(\sqrt{3}h-k)\lambda (\sqrt{3}h+k)=\left(4\sqrt{3}\lambda \right)\left(4\sqrt{3}\right)$

$3h^2-k^2=48$

Hence, the locus of (h,k) is $3x^2-y^2=48$.