First slide

Parabola

Question

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

Moderate

A

8
B

16
C

24
D

32

Solution

$Tangent at point P is t y = x + t^{2}, where the slope of$ $tangent is \tan θ = 1 / t$
$Now, the required area is$
$\begin{aligned} A = \frac{1}{2} (A N) (P N) = \frac{1}{2} (2 t^{2}) (2 t) \\ = 2 t^{3} = 2 {(t^{2})}^{3 / 2} \end{aligned}$
$Now, t^{2} \in [1, 4] . Then A_{max} occurs when t^{2} = 4 .$
$Therefore, A_{max} = 16$

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