ParabolaHardBloom L3

Question

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

Options

A.8
B.16
C.24
D.32

Solution

Sol. [Key Concept: Write the triangle's area as a function of the parameter $t$ (abscissa $= t^{2} \in \lbrack 1,4\rbrack$) and maximise it over the interval.]

Area of $\bigtriangleup PTQ = \left| \frac{1}{2} \times \left( 2t^{2} \right)(2t) \right|$

$$\Delta = 2\left| t^{3} \right|$$

$$\begin{matrix} t^{2} & \ \in \lbrack 1,4\rbrack \Rightarrow t \in \lbrack - 2, - 1\rbrack \cup \lbrack 1,2\rbrack \\ \Delta_{\max} & \ = 16\text{~}\text{for}\text{~}t = \pm 2 \end{matrix}$$

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