A tangent is drawn to the parabola y^2 = 4x a | vedclass

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(B) Let the point $P$ be $(t^2, 2t)$ on the parabola $y^2 = 4x$.The equation of the tangent at $P(t^2, 2t)$ is $ty = x + t^2$.To find the $x$-intercep

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$16$

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(B) Let the point $P$ be $(t^2, 2t)$ on the parabola $y^2 = 4x$.The equation of the tangent at $P(t^2, 2t)$ is $ty = x + t^2$.To find the $x$-intercept of the tangent,set $y = 0$,which gives $x = -t^2$. Thus,the tangent intersects the $x$-axis at $A(-t^2, 0)$.The ordinate of point $P$ is $PN = 2t$,where $N$ is the projection of $P$ on the $x$-axis,$N(t^2, 0)$.The triangle formed by the tangent,the ordinate of $P$,and the $x$-axis is $	riangle APN$.The base of the triangle is $AN = |t^2 - (-t^2)| = 2t^2$ and the height is $PN = 2t$.The area $A$ of $	riangle APN$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2t^2) 	imes (2t) = 2t^3$.Given the abscissa $t^2 \in [1, 4]$,we have $t \in [1, 2]$ (assuming $t > 0$ for the first quadrant).Thus,$A = 2t^3 = 2(t^2)^{3/2}$.Since $t^2 \in [1, 4]$,the maximum area occurs at the maximum value of $t^2$,which is $t^2 = 4$.$A_{max} = 2(4)^{3/2} = 2(8) = 16$.

$A$ tangent is drawn to the parabola $y^2 = 4x$ at the point $P(t^2, 2t)$,where the abscissa $t^2$ lies in the interval $[1, 4]$. The maximum possible area of the triangle formed by the tangent at $P$,the ordinate of the point $P$,and the $x$-axis is equal to

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