The point on the curve `sqrt(x) + sqrt(y) = sqrt(a)`, the normal at which is parallel to the x-axis, is

Options

A. (0, 0)
B. (0, a)
C. (a, 0)
D. (a, a)

Answer

To solve the problem of finding the point on the curve \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) where the normal is parallel to the x-axis, we can follow these steps:

### Step 1: Differentiate the given curve
We start with the equation of the curve:
\[ \sqrt{x} + \sqrt{y} = \sqrt{a} \]
To find the slope of the tangent line, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a}) \]
Since \(\sqrt{a}\) is a constant, its derivative is 0. Thus, we have:
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \]
Multiplying both sides by \(2\sqrt{y}\) leads to:
\[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = \frac{\sqrt{x}}{\sqrt{y}} \]

### Step 4: Set the slope of the normal to zero
Since the normal is parallel to the x-axis, its slope must be zero:
\[ \frac{\sqrt{x}}{\sqrt{y}} = 0 \]
This implies that \(\sqrt{x} = 0\), which leads to:
\[ x = 0 \]

### Step 5: Substitute \(x = 0\) into the curve equation
Now, substituting \(x = 0\) back into the original curve equation:
\[ \sqrt{0} + \sqrt{y} = \sqrt{a} \]
This simplifies to:
\[ \sqrt{y} = \sqrt{a} \]
Squaring both sides gives:
\[ y = a \]

### Final Answer
Thus, the point on the curve where the normal is parallel to the x-axis is:
\[ (0, a) \]

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