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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $y=mx$ bisects the angle between the lines
$x^2(\tan^2\theta+\cos^2\theta)+2xy\tan\theta-y^2\sin\theta=0$
when $\theta=\dfrac{\pi}{3}$, then the value of $\sqrt{3}m^2+4m$ is
Solution
For $\theta=\frac{\pi}{3}$:$\tan\theta=\sqrt3,\ \cos^2\theta=\frac14,\ \sin\theta=\frac{\sqrt3}{2}$
Equation becomes:
$x^2\left(3+\frac14\right)+2\sqrt3xy-\frac{\sqrt3}{2}y^2=0$
Angle bisector condition gives
$m=\frac{1}{\sqrt3}$
Then
$\sqrt3m^2+4m=\sqrt3\cdot\frac13+\frac{4}{\sqrt3}=\sqrt3$
Answer: $\boxed{\sqrt3}$
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Commented Jan 27 , 2026
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