A tangent is drawn to the parabola $y^2=4x$ at the point P whose abscissa lies in the interval $[1,4]$. Find the maximum area of the triangle formed by the tangent at P, ordinate of point P and the X axis.
The point P will be $(t^2,2t)$
The tangent will be
$$ty=x+t^2$$
At y=0
$$x=-t^2$$
Let the ordinate of P be PN and the tangent intersects the X axis at point A
$\Delta =\frac 12 PN. AN$
I can’t find those distances, since there is no data given. My guess is that PN will be the semi – latusrectum of the parabolla, but I am not sure.
Best Answer
We have the equation $yy_0=2(x+x_0)$ for the tangent at the point $(x_0,y_0)$. So this line intersects the $x$-axis at $x=-x_0$. $|PN|=|y_0|=2\sqrt{x_0}$ and $|AN|=2x_0$. Now by the formula for the area of the triangle $$S(x_0)=\frac{1}{2}|PN|\cdot|AN|=\frac{1}{2}\cdot 2x_0\cdot2\sqrt{x_0}=2x_0\sqrt{x_0}$$ Since $S(x_0)$ is an increasing function it attains its maximum value at the end of the segment. Hence, $$S(x_0) \le 2\cdot4\cdot\sqrt{4}=16$$ So the answer is 16.