Which among the following relations is a function ?

Options

A. `x ^(2)+y^(2)=r^(2)`
B. `(x^(2))/(a ^(2))+(y^(2))/(b ^(2))=r^(2)`
C. `y ^(2) =4ax`
D. `x ^(2) =dxy`

Answer

To determine which among the given relations is a function, we need to analyze each relation based on the definition of a function. A relation is considered a function if for every value of \( x \), there is exactly one corresponding value of \( y \).

### Step-by-Step Solution:

1. **Relation 1: \( x^2 + y^2 = r^2 \)**
- Rearranging gives \( y^2 = r^2 - x^2 \).
- Taking the square root leads to \( y = \pm \sqrt{r^2 - x^2} \).
- For each value of \( x \), there are two possible values of \( y \) (one positive and one negative).
- **Conclusion:** This relation is **not a function**.

2. **Relation 2: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2 \)**
- Rearranging gives \( \frac{y^2}{b^2} = r^2 - \frac{x^2}{a^2} \).
- This leads to \( y^2 = b^2 \left( r^2 - \frac{x^2}{a^2} \right) \).
- Taking the square root gives \( y = \pm b \sqrt{r^2 - \frac{x^2}{a^2}} \).
- Again, for each value of \( x \), there are two possible values of \( y \).
- **Conclusion:** This relation is **not a function**.

3. **Relation 3: \( y^2 = 4ax \)**
- Rearranging gives \( y = \pm \sqrt{4ax} \).
- Similar to previous cases, for each positive value of \( x \), there are two corresponding values of \( y \) (one positive and one negative).
- **Conclusion:** This relation is **not a function**.

4. **Relation 4: \( x^2 = dxy \)**
- Rearranging gives \( y = \frac{x^2}{dx} \).
- Simplifying gives \( y = \frac{x}{d} \) (assuming \( x \neq 0 \)).
- For each value of \( x \), there is exactly one corresponding value of \( y \).
- **Conclusion:** This relation **is a function**.

### Final Answer:
Among the given relations, the only relation that is a function is \( x^2 = dxy \).