Find the coordinates of the point in which the tangent at the point 'p' on the parabola x=2at, y=at^2 intersects the x-axis. Find the coordinates of the point in which the tangent at the point 'p' on the parabola x=2at, y=at^2 intersects the x-axis. I have the answer but do not know the process. THanks.
 A: HINT
the gradient of the tangent is $m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, evaluated at $t=p$
the equation of the tangent is $y-ap^2=m(x-2ap)$
it intersects the $x$ axis when $y=....?$
A: Notice, the parabola with parametric equation : $x=2at$ & $y=at^2$ can be changed  into the  cartesian form as follows $$y=a\left(\frac{x}{2a}\right)^2$$  $$x^2=4ay$$ $$\implies 4a\frac{dy}{dx}=2x$$ $$\frac{dy}{dx}=\frac{x}{2a}$$
Now, the slope of tangent at the point $P(2at, at^2)$ is given as $$m=\left[\frac{dy}{dx}\right]_{x=2at, y=at^2}=\frac{2at}{2a}=t$$ Hence the equation of the tangent at the point $P(2at, at^2)$ is given by the formula 
$$y-y_1=m(x-x_1)$$ Setting the corresponding values, we get 
$$y-at^2=t(x-2at)$$ $$y=tx-2at^2+at^2$$ $$y=tx-at^2$$
Now, at the point of intersection with the x-axis, setting $y=0$ in the equation of the tangent $$0=tx-at^2$$ $$x=\frac{at^2}{t}=at$$ Hence, the point of intersection of the tangent with the x-axis is $(at, 0)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Point of intersection}\equiv\color{blue}{(at, 0)}}}$$
