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

Option: 1

8


Option: 2

16


Option: 3

24


Option: 4

32


Answers (1)

best_answer

\mathrm{\mathrm{T}: \mathrm{ty}=\mathrm{x}+\mathrm{t}^2, \tan \theta=\frac{1}{\mathrm{t}}}

\mathrm{\mathrm{A}=\frac{1}{2}(\mathrm{AN})(\mathrm{PN})=\frac{1}{2}\left(2 \mathrm{t}^2\right)(2 \mathrm{t})}

\mathrm{A=2 t^3=2\left(t^2\right)^{3 / 2}}

\mathrm{\text { i.e. } t^2 \in[1,4] \& A_{\max } \text { occurs when } t^2=4 \Rightarrow A_{\max }=16}

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rishi.raj

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