Question

Let a line $y=mx(m>0)$ intersect the parabola $y^2=x$ at a point $P$, other than the origin. The tangent to it at $P$ meet the $x-$ axis at point $Q$. If area $△OPQ=4\text{sq.units}$, then $2m$ is equal to

Answer 1

Given parabola is $y^2=x$
Let the co-ordinates of $P$ be $(\frac{t^2}{4},\frac{t}{2})$
Putting the $P$ in $y=mx$, we get
$m=\frac{2}{t}$
Equation of tangent at $P$ is
$yt=x+\frac{t^2}{4}$
Now coordinates of $Q\equiv (-\frac{t^2}{4},0)$


Area of $△OPQ=4$
$\Rightarrow 4=\frac{1}{2}\times \text{base}\times \text{height}\Rightarrow \frac{1}{2}\times OQ\times PM=4\Rightarrow \frac{1}{2}\times \frac{t^2}{4}\times \frac{t}{2}=4\Rightarrow t=4\Rightarrow m=\frac{1}{2}\therefore 2m=1$