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. $32$

Answer 1

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 $\tan \theta =\frac{1}{t}$
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 $A=\frac{1}{2}\left(AN\right)\left(PN\right)=\frac{1}{2}\left(2t^2\right)\left(2t\right)$
$A=2t^3=2(t^2{)}^{3/2}$
Now $t^2\in [1,4]$, then $A_{max}$ occurs when $t^2=4$
$\Rightarrow A_{max}=16$