Question

The area bounded by tangent, normal and x-axis at $P(2,4)$ to the curve $y=x^2$

• A. $34$
• B. $32$
• C. $36$
• D. $24$

Answer 1

The correct option is A $34$

Slope of the tangent at $(2,4)$
$=\frac{dy}{dx}{|}_{x=2}$
$=4$.

Hence slope of normal at $(2,4)$ is $\frac{-1}{4}$.

Therefore equation of tangent will be
$y-4=4(x-2)$
$y-4=4x-8$
$4x-y=4$

Hence it cuts the x axis at $(1,0)$.

Equation of normal will be $y-4=\frac{-1}{4}(x-2)$
$4y-16=-x+2$
$x+4y=18$.

Hence it cuts the x axis at $(18,0)$.

Now the normal and tangent meet each other at $(2,4)$.

Since normal is perpendicular to the tangent it is a right angled triangle.
The coordinates are $(1,0),(18,0),(2,4)$.

Hence
Length of hypotenuse= 17.
Length of base =$\sqrt{17}$
Length of height =$\sqrt{272}=4\sqrt{17}$.

Hence
$Area=\frac{1}{2}(b\times h)$
$\frac{4\sqrt{17}\times \sqrt{17}}{2}$
$=2\times 17$
$=34$ sq units.