The equation of tangent to the curve `sqrt(x)-sqrt(y)=1` at (9,4) is

Options

A. 2x+3y=6
B. 2x+3y+6=0
C. 2x-3y+6=0
D. 2x-3y=6

Answer

To find the equation of the tangent to the curve \(\sqrt{x} - \sqrt{y} = 1\) at the point \((9, 4)\), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation:
\[ \sqrt{x} - \sqrt{y} = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(\sqrt{x}) - \frac{d}{dx}(\sqrt{y}) = 0 \]
Using the chain rule, 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 us:
\[ \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \]
Multiplying both sides by \(2\sqrt{y}\) leads to:
\[ \sqrt{y} = \sqrt{x} \cdot \frac{dy}{dx} \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at the point \((9, 4)\)
Now, substituting \(x = 9\) and \(y = 4\):
\[ \frac{dy}{dx} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3} \]

### Step 4: Use the point-slope form of the equation of the tangent line
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency. Here, \(m = \frac{2}{3}\), \(x_1 = 9\), and \(y_1 = 4\):
\[ y - 4 = \frac{2}{3}(x - 9) \]

### Step 5: Rearrange to standard form
Multiplying through by 3 to eliminate the fraction:
\[ 3(y - 4) = 2(x - 9) \]
Expanding both sides:
\[ 3y - 12 = 2x - 18 \]
Rearranging gives:
\[ 2x - 3y - 6 = 0 \]
or equivalently:
\[ 2x - 3y = 6 \]

### Final Answer
The equation of the tangent to the curve at the point \((9, 4)\) is:
\[ 2x - 3y = 6 \]