if ` y sqrt(1-x^2) = k - x sqrt(1-y^2)` and `y(1/2) = -1/4`, then `(dy)/(dx) ` at x = ` 1/2 `

Options

A. (A) `sqrt5/2`
B. (B) `-sqrt5/4`
C. (C) `2/sqrt5`
D. (D) `-sqrt5/2`

Explanation

To solve the problem, we need to differentiate the given equation and find the value of \(\frac{dy}{dx}\) at \(x = \frac{1}{2}\) when \(y\left(\frac{1}{2}\right) = -\frac{1}{4}\).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ y \sqrt{1 - x^2} = k - x \sqrt{1 - y^2} \]

2. **Rearrange the equation:**
\[ y \sqrt{1 - x^2} + x \sqrt{1 - y^2} - k = 0 \]

3. **Differentiate both sides with respect to \(x\):**
Using implicit differentiation:
\[ \frac{d}{dx}\left(y \sqrt{1 - x^2}\right) + \frac{d}{dx}\left(x \sqrt{1 - y^2}\right) - \frac{dk}{dx} = 0 \]
Since \(k\) is a constant, \(\frac{dk}{dx} = 0\).

4. **Apply the product rule:**
For the first term:
\[ \frac{d}{dx}\left(y \sqrt{1 - x^2}\right) = \sqrt{1 - x^2} \frac{dy}{dx} + y \cdot \frac{d}{dx}\left(\sqrt{1 - x^2}\right) \]
\[ = \sqrt{1 - x^2} \frac{dy}{dx} - \frac{yx}{\sqrt{1 - x^2}} \]

For the second term:
\[ \frac{d}{dx}\left(x \sqrt{1 - y^2}\right) = \sqrt{1 - y^2} + x \cdot \frac{d}{dx}\left(\sqrt{1 - y^2}\right) \]
\[ = \sqrt{1 - y^2} - \frac{xy}{\sqrt{1 - y^2}} \frac{dy}{dx} \]

5. **Combine the derivatives:**
Putting it all together:
\[ \sqrt{1 - x^2} \frac{dy}{dx} - \frac{yx}{\sqrt{1 - x^2}} + \sqrt{1 - y^2} - \frac{xy}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \]

6. **Group the \(\frac{dy}{dx}\) terms:**
\[ \left(\sqrt{1 - x^2} - \frac{xy}{\sqrt{1 - y^2}}\right) \frac{dy}{dx} = \frac{yx}{\sqrt{1 - x^2}} - \sqrt{1 - y^2} \]

7. **Solve for \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = \frac{\frac{yx}{\sqrt{1 - x^2}} - \sqrt{1 - y^2}}{\sqrt{1 - x^2} - \frac{xy}{\sqrt{1 - y^2}}} \]

8. **Substitute \(x = \frac{1}{2}\) and \(y = -\frac{1}{4}\):**
Calculate the necessary values:
\[ \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
\[ \sqrt{1 - \left(-\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]

Substitute into the equation:
\[ \frac{dy}{dx} = \frac{\left(-\frac{1}{4} \cdot \frac{1}{2}\right) \cdot \frac{2}{\sqrt{3}} - \frac{\sqrt{15}}{4}}{\frac{\sqrt{3}}{2} - \left(-\frac{1}{4} \cdot \frac{\sqrt{15}}{4}\right)} \]

9. **Simplify the expression:**
After substituting and simplifying, we can find the final value of \(\frac{dy}{dx}\).

### Final Result:
\[ \frac{dy}{dx} = -\frac{\sqrt{5}}{2} \]