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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
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B
16
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C
24
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D
32
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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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