# Let a line y=m x(m>0) intersect the parabola

Question:

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

Solution:

Let the coordinates of $P=P\left(t^{2}, t\right)$

Tangent at $P\left(t^{2}, t\right)$ is $t y=\frac{x+t^{2}}{2}$

$\Rightarrow \quad 2 t y=x+t^{2}$

$Q\left(-t^{2}, 0\right), O(0,0)$

$\therefore \quad$ Area of $\Delta O P Q=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ t^{2} & t & 1 \\ -t^{2} & 0 & 1\end{array}\right|=4$

$\Rightarrow|t|^{3}=8$

$t=\pm 2(t>0)$

$\therefore \quad 4 y=x+4$ is a tangent

$\therefore \quad P$ is $(4,2)$

Now, $y=m x \quad \therefore m=\frac{1}{2}$