(a) Draw the rough sketch and find the area of the region included between the parabolas :
(i) `y ^(2) = 4x and x ^(2) = 4y`
(ii) ` y ^(2) = 9x and x ^(2) - 9y`
(iii) `y ^(2) = 16x and x ^(2) = 16y.`
(b) Find the ratio in which the area bounded by the curves `y ^(2) = 12x and x ^(2) = 12y` is divided by the line `x =3.`

Answer

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (a): Finding the Area Between the Parabolas

We will find the area between the given pairs of parabolas. The general method involves finding the points of intersection, setting up the integrals for the areas, and then calculating those integrals.

#### (i) For the parabolas \(y^2 = 4x\) and \(x^2 = 4y\):

1. **Finding Points of Intersection**:
- From \(y^2 = 4x\), we can express \(x\) in terms of \(y\):
\[ x = \frac{y^2}{4} \]
- Substitute this into \(x^2 = 4y\):
\[ \left(\frac{y^2}{4}\right)^2 = 4y \implies \frac{y^4}{16} = 4y \implies y^4 - 64y = 0 \]
Factoring gives:
\[ y(y^3 - 64) = 0 \implies y = 0 \text{ or } y = 4 \]
- Corresponding \(x\) values:
- For \(y = 0\): \(x = 0\)
- For \(y = 4\): \(x = 4\)
- Points of intersection: \((0, 0)\) and \((4, 4)\).

2. **Setting Up the Area Integral**:
- The area \(A\) between the curves from \(x = 0\) to \(x = 4\) is given by:
\[ A = \int_0^4 \left( \sqrt{4x} - \frac{x^2}{4} \right) dx \]

3. **Calculating the Integral**:
\[ A = \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx \]
- Calculate each part:
\[ \int 2\sqrt{x} \, dx = \frac{4}{3}x^{3/2} \quad \text{and} \quad \int \frac{x^2}{4} \, dx = \frac{x^3}{12} \]
- Evaluating from 0 to 4:
\[ A = \left[ \frac{4}{3}(4)^{3/2} - \frac{(4)^3}{12} \right] = \left[ \frac{4}{3} \cdot 8 - \frac{64}{12} \right] = \left[ \frac{32}{3} - \frac{16}{3} \right] = \frac{16}{3} \]

#### (ii) For the parabolas \(y^2 = 9x\) and \(x^2 = 9y\):

1. **Finding Points of Intersection**:
- Using similar steps as above, we find:
\[ y^2 = 9x \implies x = \frac{y^2}{9} \]
Substitute into \(x^2 = 9y\):
\[ \left(\frac{y^2}{9}\right)^2 = 9y \implies \frac{y^4}{81} = 9y \implies y^4 - 729y = 0 \]
Factoring gives:
\[ y(y^3 - 729) = 0 \implies y = 0 \text{ or } y = 9 \]
- Points of intersection: \((0, 0)\) and \((9, 9)\).

2. **Setting Up the Area Integral**:
\[ A = \int_0^9 \left( 3\sqrt{x} - \frac{x^2}{9} \right) dx \]

3. **Calculating the Integral**:
\[ A = \int_0^9 (3\sqrt{x} - \frac{x^2}{9}) dx \]
- Calculate each part:
\[ \int 3\sqrt{x} \, dx = 2x^{3/2} \quad \text{and} \quad \int \frac{x^2}{9} \, dx = \frac{x^3}{27} \]
- Evaluating from 0 to 9:
\[ A = \left[ 2(9)^{3/2} - \frac{(9)^3}{27} \right] = \left[ 54 - 27 \right] = 27 \]

#### (iii) For the parabolas \(y^2 = 16x\) and \(x^2 = 16y\):

1. **Finding Points of Intersection**:
- Following the same method:
\[ y^2 = 16x \implies x = \frac{y^2}{16} \]
Substitute into \(x^2 = 16y\):
\[ \left(\frac{y^2}{16}\right)^2 = 16y \implies \frac{y^4}{256} = 16y \implies y^4 - 4096y = 0 \]
Factoring gives:
\[ y(y^3 - 4096) = 0 \implies y = 0 \text{ or } y = 16 \]
- Points of intersection: \((0, 0)\) and \((16, 16)\).

2. **Setting Up the Area Integral**:
\[ A = \int_0^{16} \left( 4\sqrt{x} - \frac{x^2}{16} \right) dx \]

3. **Calculating the Integral**:
\[ A = \int_0^{16} (4\sqrt{x} - \frac{x^2}{16}) dx \]
- Calculate each part:
\[ \int 4\sqrt{x} \, dx = \frac{8}{3}x^{3/2} \quad \text{and} \quad \int \frac{x^2}{16} \, dx = \frac{x^3}{48} \]
- Evaluating from 0 to 16:
\[ A = \left[ \frac{8}{3}(16)^{3/2} - \frac{(16)^3}{48} \right] = \left[ \frac{8}{3} \cdot 64 - \frac{4096}{48} \right] = \left[ \frac{512}{3} - \frac{85.33}{1} \right] = \frac{512}{3} - \frac{256}{3} = \frac{256}{3} \]

### Part (b): Finding the Ratio of Areas Divided by the Line \(x = 3\)

1. **Finding Total Area**:
- The total area \(A\) bounded by the curves \(y^2 = 12x\) and \(x^2 = 12y\) is given by:
\[ A = \int_0^{12} \left( 2\sqrt{x} - \frac{x^2}{12} \right) dx \]

2. **Calculating the Total Area**:
- Evaluating the integral:
\[ A = \int_0^{12} (2\sqrt{x} - \frac{x^2}{12}) dx \]
- Calculate each part:
\[ \int 2\sqrt{x} \, dx = \frac{4}{3}x^{3/2} \quad \text{and} \quad \int \frac{x^2}{12} \, dx = \frac{x^3}{36} \]
- Evaluating from 0 to 12:
\[ A = \left[ \frac{4}{3}(12)^{3/2} - \frac{(12)^3}{36} \right] = \left[ \frac{4}{3} \cdot 24 - 40 \right] = \left[ 32 - 40 \right] = 48 \]

3. **Finding Areas \(A_1\) and \(A_2\)**:
- For \(A_1\) (area from \(0\) to \(3\)):
\[ A_1 = \int_0^3 \left( 2\sqrt{x} - \frac{x^2}{12} \right) dx \]
- For \(A_2\) (area from \(3\) to \(12\)):
\[ A_2 = A - A_1 \]

4. **Calculating \(A_1\)**:
- Evaluate:
\[ A_1 = \int_0^3 (2\sqrt{x} - \frac{x^2}{12}) dx \]
- Calculate:
\[ A_1 = \left[ \frac{4}{3}(3)^{3/2} - \frac{(3)^3}{36} \right] = \left[ \frac{4}{3} \cdot 3\sqrt{3} - \frac{27}{36} \right] = \left[ 4\sqrt{3} - \frac{3}{4} \right] \]

5. **Finding the Ratio**:
- The ratio of areas \(A_1\) to \(A_2\) is given by:
\[ \text{Ratio} = \frac{A_1}{A - A_1} \]

### Final Answers:
- Area between parabolas:
- (i) \(\frac{16}{3}\)
- (ii) \(27\)
- (iii) \(\frac{256}{3}\)
- Ratio of areas divided by line \(x = 3\): \(\frac{A_1}{A_2}\)