For what value of `'y', x^(2)=(1)/(12)x-y^(2)` is a perfect square ?

Options

A. `1//24`
B. `1//12`
C. `1//6`
D. `1//3`

Answer

To solve the problem, we want to find the value of \( y \) such that the expression \( x^2 = \frac{1}{12}x - y^2 \) is a perfect square.

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start with the given equation:
\[ x^2 = \frac{1}{12}x - y^2 \]
Rearranging it gives:
\[ x^2 + y^2 - \frac{1}{12}x = 0 \]

2. **Reorganizing the Terms**:
We can reorganize the equation as:
\[ x^2 - \frac{1}{12}x + y^2 = 0 \]

3. **Identifying the Form**:
Notice that the equation resembles the form of a perfect square trinomial:
\[ a^2 - 2ab + b^2 = (a - b)^2 \]
Here, we can let \( a = x \) and \( b = y \).

4. **Comparing with the Perfect Square Form**:
We can express the equation in the form:
\[ (x - b)^2 = b^2 - y^2 \]
By comparing, we identify:
\[ -2ab = -\frac{1}{12}x \]
This gives us:
\[ 2xy = \frac{1}{12}x \]

5. **Solving for \( y \)**:
Dividing both sides by \( x \) (assuming \( x \neq 0 \)):
\[ 2y = \frac{1}{12} \]
Thus, solving for \( y \):
\[ y = \frac{1}{24} \]

6. **Conclusion**:
Therefore, the value of \( y \) for which \( x^2 = \frac{1}{12}x - y^2 \) is a perfect square is:
\[ \boxed{\frac{1}{24}} \]