Question

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

A
8
B
16
C
24
D
32

Solution

## The correct option is B $$16$$Equation of parabola is$$y^2 =4x$$Here $$a=1$$.Let $$P(t^2,2t)$$ be any point on the parabola.Equation of tangent at $$P$$ is $$ty = x + t^2$$, where slope of tangent is $$\tan \theta = \dfrac{1}{t}$$Since the tangent passes through $$x-$$axis i.e.$$y=0$$.So, $$x=-t^2$$.So, $$A(-t^2,0)$$ is the point of intersection of tangent and x-axis.Now required area is $$A = \dfrac{1}{2} (AN) (PN) = \dfrac{1}{2} (2t^2)(2t)$$$$A = 2t^3 = 2(t^2)^{3/2}$$Now $$t^2 \in [1, 4]$$, then  $$A_{max}$$ occurs when $$t^2 = 4$$$$\Rightarrow A_{max} = 16$$Mathematics

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