#### To determine

**To find:**

Equation for C

#### Answer

y=329x2

#### Explanation

**1) Concept:**

Construct integrals for area A and B, and then find the equation of curve C.

**2) Calculation:**

As point P lies on the curve y=2x2, with x=a, the y co-ordinate of point P is 2a2.

So, let coordinates of point P be (a, 2a2).

To calculate the area A:

It is the area between curve y=2x2 and y=x2, between x=0 and x=a. So

A=∫0a2x2-x2dx

A=∫0ax2dx

A=x330a

A=a33

To calculate area B: It is the area between curves C and y=2x2 where y lies between 0 and a2.

It lies in the first quadrant.

Solve y=2x2 for x.

So,

x= fy=y2

Consider the equation for curve C as x=C(y).

So the area between curves is given by,

B=∫02a2y2-C(y)dy

B=43y23202a2-∫02a2Cydy

B=43a3-∫02a2Cydy

Given that areas are equal,

So, A=B

a33=43a3-∫02a2Cydy

∫02a2Cydy=4a33-a33

∫02a2Cydy=a3

Differentiating with respect to a,

dda∫02a2Cydy=3a2

By the first fundamental theorem of calculus and chain rule,

C2a24a=3a2

C2a2=3a24a

C2a2=3a4

Now, let y=2a2

Cy=3a4

But, a=y2. So,

Cy=34·y2

As Cy=x

x=34·y2

Solving the equation for y:

By squaring,

x2=916·y2

y=329x2

This is the required equation for C

**Conclusion:**

y=329x2