Let a line y=mx (m0) intersect the parabola | vedclass

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(B) The parabola is $y^2=x$. Let $P$ be $(t^2, t)$ for $t>0$.The line $y=mx$ passes through $P(t^2, t)$,so $t=m(t^2)$,which gives $m=1/t$.The tangent

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$0.5$

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(B) The parabola is $y^2=x$. Let $P$ be $(t^2, t)$ for $t>0$.The line $y=mx$ passes through $P(t^2, t)$,so $t=m(t^2)$,which gives $m=1/t$.The tangent to $y^2=x$ at $P(t^2, t)$ is $ty = \frac{1}{2}(x+t^2)$.Setting $y=0$ to find the $x$-intercept $Q$,we get $0 = \frac{1}{2}(x+t^2)$,so $x = -t^2$. Thus,$Q$ is $(-t^2, 0)$.The vertices of $\Delta OPQ$ are $O(0, 0)$,$P(t^2, t)$,and $Q(-t^2, 0)$.The area is $\frac{1}{2} |x_O(y_P-y_Q) + x_P(y_Q-y_O) + x_Q(y_O-y_P)| = 4$.$\frac{1}{2} |0(t-0) + t^2(0-0) + (-t^2)(0-t)| = 4$.$\frac{1}{2} |t^3| = 4 \implies t^3 = 8 \implies t = 2$.Since $m = 1/t$,we have $m = 1/2 = 0.5$.

Let a line $y=mx$ $(m>0)$ 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 $\text{area}(\Delta OPQ)=4$ sq. units,then $m$ is equal to

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