The sum of the intercepts made on the axes of coordinates by any tangent to the curve ` sqrt(x)+sqrt(y)=sqrt(a)` is equal to

Options

A. 2a
B. a
C. `(a)/(2)`
D. `2sqrt(a)`

Answer

To solve the problem, we need to find the sum of the intercepts made on the coordinate axes by any tangent to the curve given by the equation \( \sqrt{x} + \sqrt{y} = \sqrt{a} \).

### Step-by-Step Solution:

1. **Understanding the Curve**:
The curve is defined by the equation \( \sqrt{x} + \sqrt{y} = \sqrt{a} \). This represents a curve in the first quadrant of the coordinate system.

2. **Finding the Derivative**:
To find the slope of the tangent at any point on the curve, we differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a}) \]
This gives:
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \]
Rearranging this, we find:
\[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]

3. **Equation of the Tangent Line**:
The equation of the tangent line at a point \( (x_1, y_1) \) on the curve can be written using the point-slope form:
\[ y - y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}}(x - x_1) \]

4. **Finding the x-intercept**:
To find the x-intercept (where \( y = 0 \)):
\[ 0 - y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}}(x - x_1) \]
Rearranging gives:
\[ y_1 = \frac{\sqrt{y_1}}{\sqrt{x_1}}(x - x_1) \]
Solving for \( x \):
\[ x = x_1 + \frac{y_1 \sqrt{x_1}}{\sqrt{y_1}} = x_1 + \sqrt{x_1 y_1} \]
Thus, the x-intercept \( OA = x_1 + \sqrt{x_1 y_1} \).

5. **Finding the y-intercept**:
To find the y-intercept (where \( x = 0 \)):
\[ y - y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}}(0 - x_1) \]
Rearranging gives:
\[ y = y_1 + \frac{\sqrt{y_1}}{\sqrt{x_1}} x_1 = y_1 + \sqrt{x_1 y_1} \]
Thus, the y-intercept \( OB = y_1 + \sqrt{x_1 y_1} \).

6. **Sum of the Intercepts**:
Now we can find the sum of the intercepts:
\[ OA + OB = (x_1 + \sqrt{x_1 y_1}) + (y_1 + \sqrt{x_1 y_1}) = x_1 + y_1 + 2\sqrt{x_1 y_1} \]

7. **Using the Curve Equation**:
From the curve equation \( \sqrt{x_1} + \sqrt{y_1} = \sqrt{a} \), we can express \( x_1 + y_1 \) in terms of \( a \):
\[ x_1 + y_1 = (\sqrt{x_1} + \sqrt{y_1})^2 - 2\sqrt{x_1 y_1} = a - 2\sqrt{x_1 y_1} \]

8. **Final Result**:
Substituting back into the sum of intercepts:
\[ OA + OB = a \]

Thus, the sum of the intercepts made on the axes of coordinates by any tangent to the curve \( \sqrt{x} + \sqrt{y} = \sqrt{a} \) is equal to \( a \).