question_answer

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 point 'P' and the x-axis is equal to -

A) 8                                 

B) 16   

C) 24                                

D) 32

Correct Answer: B

Solution :

\[{{y}^{2}}=4x\]	??..(1)
Equation of tangent at point
P is \[\Rightarrow ty=x+{{y}^{2}}\]
where, slope of tangent is
\[\tan \theta =\frac{1}{t}\]

Now, required area is
\[\Delta =\frac{1}{2}(AN)\,\,(PN)\]
\[=\frac{1}{2}(2{{t}^{2}})\,\,(2t)\]
\[\Delta =2{{t}^{3}}=2{{({{t}^{2}})}^{3/2}}\]	..? (2)
Now, \[{{t}^{2}}\in [1,4]\]
then \[{{\Delta }_{\max imum}}=2{{(4)}^{3/2}}=16,\]when \[{{t}^{2}}=4\]