Question

If the area enclosed by ${{y}^{2}}=4ax$ and the line y = ax is $\dfrac{1}{3}$ square units, then the area enclosed by $y=4x$ with same parabola is
A . 8 square units
B. 4 square units
C. $\dfrac{4}{3}$square units
D. $\dfrac{8}{3}$ square units

Answer 1

Hint:In this question, we have to find the area under curve. First we have to find out the value of a. for this we have to find the intersection points and then apply the limits of integration then after finding it, we again apply the formula of area under curve for the given curve with the same parabola to find the desired answer.

Formula used:
Formula for area under curve is = $\int_{a}^{b}{f(x)dx}$ where (b > a)

Complete step by step Solution:
Here we are given the area enclosed by ${{y}^{2}}=4ax$and the line y = ax is $\dfrac{1}{3}$square units
Here ${{y}^{2}}=4ax$then y = $\sqrt{4ax}$ and y = ax
First we find the point of intersection which are given as
$ax=\sqrt{4ax}$
Squaring both sides, we get
${{a}^{2}}{{x}^{2}}=4ax$
That is ${{a}^{2}}{{x}^{2}}-4ax=0$
Or $ax(ax-4)=0$
That is ax = 0 or ax – 4 = 0
Then x = 0 or $x=\dfrac{4}{a}$
When x = 0 from y = ax
Then y = 0
When $x=\dfrac{4}{a}$ then from y = ax
Then $y=a\times \dfrac{4}{a}$ = 4

Thus the limit of integration is y = 0 to y = 4
Therefore, area of the shaded region = $\int_{0}^{4}{\left( \dfrac{y}{a} \right)}-{{\left( \dfrac{y}{4a} \right)}^{2}}dx$
Now we integrate the above equation and we get
= $2,-4$${{\left[ \dfrac{{{y}^{2}}}{2a}-\dfrac{{{y}^{3}}}{12a} \right]}_{0}}^{4}$
By solving it, we get $\left[ \dfrac{{{(4)}^{2}}}{2a}-\dfrac{{{(4)}^{3}}}{12a}-\left( \dfrac{{{(0)}^{2}}}{2a}-\dfrac{{{(0)}^{3}}}{12a} \right) \right]$
That is $\left[ \dfrac{16}{2a}-\dfrac{64}{12a} \right]$ = $\dfrac{8}{3a}$
And $\dfrac{8}{3a}$= $\dfrac{1}{3}$
Then $a=8$
Intersection point of $y=4x$ and ${{y}^{2}}=4ax$ is given by :
$16{{x}^{2}}=32x$
Then $16x(x-2)=0$
This means $x=0,y=0$ and $x=2,y=8$
Area enclosed by $y=4x$ and ${{y}^{2}}=4ax$ is given by :
$A=\int_{0}^{8}{\left( \dfrac{y}{4}-\dfrac{{{y}^{2}}}{32} \right)}dx$
Integrating the above equation, we get
$A={{\left[ \dfrac{{{y}^{2}}}{8}-\dfrac{{{y}^{3}}}{32\times 3} \right]}_{3}}^{8}$
Then $A=\dfrac{{{8}^{2}}}{8}-\dfrac{{{8}^{3}}}{32\times 3}$
So $A=\dfrac{8}{3}$

Therefore, the correct option is (D).

Note: To find the area we can also use another method which is : first we can find the area enclosed by parabola from A to B then we find the area of triangle made by line y = ax from A to B. After finding the difference of both the areas we find the required enclosed area