A tangent is drawn to the parabola `y^2=4a x` at the point `P` whose abscissa lies in the interval (1, 4). The maximum possible area f the triangle formed by the tangent at `P ,` the ordinates of the point `P ,` and the x-axis is equal to 8 (b) 16 (c) 24 (d) 32

Options

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

Answer

B

Explanation

(2) Tangent at point P is `ty=x+t^(2)`, where the slope of tangent is `tantheta=1//t`.
Now, the required area is
`A=(1)/(2)(AN)(PN)`
`=(1)/(2)(2t^(2))(2t)`
`=2t^(3)=2(t^(2))^(3//2)`
Now, `t^(2)in[1,4]." Then "A_(max)`
occurs when `t^(2)=4`.
Therefore, `A_(max)=16`.