If the area enclosed between the curves `y=kx^2` and `x=ky^2`, where `kgt0`, is 1 square unit. Then k is: (a) `1/sqrt(3)` (b) `sqrt(3)/2` (c) `2/sqrt(3)` (d) `sqrt(3)`

Options

A. `1/sqrt3`
B. `2/sqrt3`
C. `sqrt3/2`
D. `sqrt3`

Answer

To solve the problem, we need to find the value of \( k \) such that the area enclosed between the curves \( y = kx^2 \) and \( x = ky^2 \) is equal to 1 square unit. Let's go through the steps systematically.

### Step 1: Set up the equations
We have two curves:
1. \( y = kx^2 \)
2. \( x = ky^2 \)

From the second equation, we can express \( y \) in terms of \( x \):
\[ y = \sqrt{\frac{x}{k}} \]

### Step 2: Find the points of intersection
To find the points where these curves intersect, we set the two equations equal to each other:
\[ kx^2 = \sqrt{\frac{x}{k}} \]
Squaring both sides to eliminate the square root gives:
\[ k^2x^4 = \frac{x}{k} \]
Multiplying both sides by \( k \) (since \( k > 0 \)):
\[ k^3x^4 = x \]
Rearranging this, we get:
\[ k^3x^4 - x = 0 \]
Factoring out \( x \):
\[ x(k^3x^3 - 1) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( k^3x^3 = 1 \) which leads to \( x = \frac{1}{k} \)

### Step 3: Find the corresponding y-values
Now we find the corresponding \( y \) values for the intersection points:
- For \( x = 0 \):
\[ y = k(0)^2 = 0 \]
- For \( x = \frac{1}{k} \):
\[ y = k\left(\frac{1}{k}\right)^2 = \frac{1}{k} \]

So the points of intersection are \( (0, 0) \) and \( \left(\frac{1}{k}, \frac{1}{k}\right) \).

### Step 4: Calculate the area between the curves
The area \( A \) enclosed between the curves from \( x = 0 \) to \( x = \frac{1}{k} \) can be calculated using the integral:
\[ A = \int_0^{\frac{1}{k}} \left( \sqrt{\frac{x}{k}} - kx^2 \right) dx \]

### Step 5: Solve the integral
Calculating the integral:
\[ A = \int_0^{\frac{1}{k}} \sqrt{\frac{x}{k}} \, dx - \int_0^{\frac{1}{k}} kx^2 \, dx \]

1. The first integral:
\[ \int_0^{\frac{1}{k}} \sqrt{\frac{x}{k}} \, dx = \sqrt{\frac{1}{k}} \int_0^{\frac{1}{k}} x^{1/2} \, dx = \sqrt{\frac{1}{k}} \left[ \frac{2}{3} x^{3/2} \right]_0^{\frac{1}{k}} = \sqrt{\frac{1}{k}} \cdot \frac{2}{3} \left( \frac{1}{k} \right)^{3/2} = \frac{2}{3k^{2}} \]

2. The second integral:
\[ \int_0^{\frac{1}{k}} kx^2 \, dx = k \left[ \frac{x^3}{3} \right]_0^{\frac{1}{k}} = k \cdot \frac{1}{3k^3} = \frac{1}{3k^2} \]

### Step 6: Combine the results
Now, substituting back into the area equation:
\[ A = \frac{2}{3k^2} - \frac{1}{3k^2} = \frac{1}{3k^2} \]

### Step 7: Set the area equal to 1
We set the area equal to 1 square unit:
\[ \frac{1}{3k^2} = 1 \]
This simplifies to:
\[ 3k^2 = 1 \implies k^2 = \frac{1}{3} \implies k = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]