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,ordinate of point P and the x-axis is equal to?
(ans=16)

Answer 1

Answer:

Equation of parabola is

y2=4x

Here a=1.

Let P(t2,2t) be any point on the parabola.

Equation of tangent at P is ty=x+t2, where slope of tangent is tanθ=t1

Since the tangent passes through x−axis i.e.y=0.So, x=−t2.

So, A(−t2,0) is the point of intersection of tangent and x-axis.

Now required area is A=21(AN)(PN)=21(2t2)(2t)

A=2t3=2(t2)3/2

Now t2∈[1,4], then  Amax occurs when t2=4

⇒Amax=16