The slope of the tangent to the curve `y=sqrt(4-x^2)` at the point where the ordinate and the abscissa are equal is (a) `-1` (b) 1 (c) 0 (d) none of these

Options

A. `-1`
B. 1
C. 0
D. none of these

Answer

To find the slope of the tangent to the curve \( y = \sqrt{4 - x^2} \) at the point where the ordinate and the abscissa are equal, we can follow these steps:

### Step 1: Identify the condition where ordinate and abscissa are equal
Since the ordinate (y-coordinate) and the abscissa (x-coordinate) are equal, we can set \( x = y \).

### Step 2: Substitute \( y \) in the equation
From the equation of the curve, we have:
\[ y = \sqrt{4 - x^2} \]
Setting \( x = y \), we can substitute \( y \) into the equation:
\[ x = \sqrt{4 - x^2} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ x^2 = 4 - x^2 \]

### Step 4: Rearranging the equation
Rearranging the equation, we get:
\[ x^2 + x^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2} \text{ or } x = -\sqrt{2} \]

### Step 5: Find the corresponding \( y \) values
Since \( y = x \), we have:
\[ y = \sqrt{2} \text{ or } y = -\sqrt{2} \]
Thus, the points of interest are \( (\sqrt{2}, \sqrt{2}) \) and \( (-\sqrt{2}, -\sqrt{2}) \).

### Step 6: Differentiate the function to find the slope
Now, we need to find the derivative \( \frac{dy}{dx} \) to determine the slope of the tangent line. The function is:
\[ y = \sqrt{4 - x^2} \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{4 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{4 - x^2}} \]

### Step 7: Evaluate the derivative at the point \( x = \sqrt{2} \)
Substituting \( x = \sqrt{2} \) into the derivative:
\[ \frac{dy}{dx} = \frac{-\sqrt{2}}{\sqrt{4 - (\sqrt{2})^2}} = \frac{-\sqrt{2}}{\sqrt{4 - 2}} = \frac{-\sqrt{2}}{\sqrt{2}} = -1 \]

### Conclusion
The slope of the tangent to the curve at the point where the ordinate and the abscissa are equal is \( -1 \).

### Final Answer
Thus, the correct option is (a) \( -1 \).

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