Question

Area bounded by the 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

Hint:
Here we have to find the area bounded by the tangent, normal and x-axis at P (2, 4) to the curve $y={{x}^{2}}$. For that, we will first calculate the slope of the tangent of the curve at point P (2, 4), then we will find the equation of the tangent. Similarly we will find the slope of the normal at point P (2, 4) and then we will find the equation of the normal. Then we will calculate the points where the normal and tangent of the curve cut the x-axis and then we will calculate the intersection point of the tangent and the normal. All these three points will form a right triangle and we can easily calculate the base and height of the triangle, and then we will calculate the area of this right angled triangle which is the required area which we needed.

Complete step by step solution:
Here the given curve is $y={{x}^{2}}$
We will calculate the slope of the tangent at P (2, 4), which is equal to
Slope $={{\left. \dfrac{dy}{dx} \right|}_{x=2}}$
Putting value of y here
Slope $={{\left. \dfrac{d{{x}^{2}}}{dx} \right|}_{x=2}}$
Differentiating the term with respect to x, we get
Slope $={{\left. 2x \right|}_{x=2}}$
Now, we will put the value of x here
Slope $=2\times 2=4$
Therefore, the slope of the tangent is 4
Now, we will write the equation of the tangent of slope 4.
$\Rightarrow y-4=4(x-2)$
Further simplifying the equation, we get
$\Rightarrow y-4x+4=0$……………….$\left( 1 \right)$
Similarly, We will calculate the slope of the normal at P (2, 4), which is equal to
Slope $=-\dfrac{1}{{{\left. \dfrac{dy}{dx} \right|}_{x=2}}}$
Putting value of y here
Slope $=-\dfrac{1}{{{\left. \dfrac{d{{x}^{2}}}{dx} \right|}_{x=2}}}$
Differentiating the term with respect to x, we get
Slope $={{\left. -\dfrac{1}{2x} \right|}_{x=2}}$
Now, we will put the value of x here
Slope $=-\dfrac{1}{2\times 2}=-\dfrac{1}{4}$
Therefore, the slope of the normal is $-\dfrac{1}{4}$
Now, we will write the equation of the normal of slope $-\dfrac{1}{4}$
$y-4=-\dfrac{1}{4}(x-2)$
Further simplifying the equation, we get
$\Rightarrow x+4y=18$…………………….$\left( 2 \right)$
Now, we will find the point on the x-axis, where the tangent cuts the x-axis by putting y=0 in the equation of the tangent.
$\Rightarrow 0-4x+4=0$
Simplifying the terms, we get
$\Rightarrow x=1$
Therefore, the point is $(1,0)$
Similarly, we will find the point on the x-axis, where the normal cuts the x-axis by putting y=0 in the equation of the normal.
$\Rightarrow x+4\times 0=18$
Simplifying the terms, we get
$\Rightarrow x=18$
Therefore, the point is $(18,0)$
Now, we will find the point of intersection of the tangent and the normal by solving the equation of tangent and normal.
For that, we will multiply equation 2 with 4 and then we will add it to equation 1.
$\Rightarrow 4x+16y+y-4x+4=52$
Simplifying the terms, we get
$\begin{align}
  &\Rightarrow 17y=68 \\ \Rightarrow
 & y=4 \\
\end{align}$
Putting value of y in equation 1, we get
$\begin{align}
  &\Rightarrow 4-4x+4=0 \\ \Rightarrow
 & x=2 \\
\end{align}$
Therefore, the point of intersection of the tangent and the normal is (2, 4).
As tangent and normal intersect each other perpendicularly, so these three points will form a right angled triangle. The points are $\left( 1,0 \right),\left( 18,0 \right)\And \left( 2,4 \right)$
Now, we will calculate the distance between all three points one by one using distance formula.
Distance of the point $(1,0)$ from $(18,0)$ $=\sqrt{\left[ {{\left( 18-1 \right)}^{2}}+{{\left( 0-0 \right)}^{2}} \right]}$
We will subtract 1 from 18 and then we will take the square of it and then square root.
 Distance of the point $(1,0)$ from $(18,0)$ $=17$
Distance of the point $(2,4)$ from $(18,0)$ $=\sqrt{\left[ {{\left( 18-2 \right)}^{2}}+{{\left( 0-4 \right)}^{2}} \right]}$
We will calculate the difference of the terms inside the bracket.
Distance of the point $(2,4)$ from $(18,0)$ $=\sqrt{\left[ {{16}^{2}}+{{4}^{2}} \right]}$
Now, we will calculate the square of these two terms and then we will add both the terms.



Distance of the point $(2,4)$from $(1,0)$ $=\sqrt{\left[ {{\left( 1-2 \right)}^{2}}+{{\left( 0-4 \right)}^{2}} \right]}$
We will calculate the difference of the terms inside the bracket.
Distance of the point $(2,4)$from $(1,0)$ $=\sqrt{\left[ {{1}^{2}}+{{4}^{2}} \right]}$
Now, we will calculate the square of these two terms and then we will add both the terms.
Distance of the point $(2,4)$from $(1,0)$ $=\sqrt{17}$

Here the hypotenuse is 17 as this is the longest side.
Base of the triangle $=4\sqrt{17}$
Height of the triangle $=\sqrt{17}$
Area of the triangle $=\dfrac{1}{2}\times base\times height=\dfrac{1}{2}\times 4\sqrt{17}\times \sqrt{17}$
$=34$

Hence, area bounded by the tangent, normal and x-axis at P (2, 4) to the curve $y={{x}^{2}}$ $=34$
Thus, the correct option is A.

Note:
The formula which we have used here for finding the distance between the two points is in two dimensional.
The distance formula for three dimensional is different.
Say we have to find the distance between the points $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and $({{x}_{2}},{{y}_{2}},{{z}_{2}})$
So the distance between these two points is
$\sqrt{\left[ {{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}} \right]}$