Find the points on the curve `y=sqrt(x-3)`, where the tangent is perpendicular to the line `6x+3y-5=0`.

Answer

To find the points on the curve \( y = \sqrt{x - 3} \) where the tangent is perpendicular to the line \( 6x + 3y - 5 = 0 \), we will follow these steps:

### Step 1: Find the slope of the line
First, we need to rewrite the line equation in slope-intercept form \( y = mx + c \).

Starting with the equation:
\[ 6x + 3y - 5 = 0 \]
Rearranging gives:
\[ 3y = -6x + 5 \]
Dividing through by 3:
\[ y = -2x + \frac{5}{3} \]

From this, we can see that the slope \( m \) of the line is \( -2 \).

### Step 2: Find the slope of the curve
Next, we need to find the derivative of the curve \( y = \sqrt{x - 3} \) to determine the slope of the tangent at any point on the curve.

Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2\sqrt{x - 3}} \]

### Step 3: Set up the condition for perpendicularity
For the tangent to be perpendicular to the line, the product of their slopes must equal \(-1\). Thus, we set up the equation:
\[ \left(\frac{1}{2\sqrt{x - 3}}\right) \cdot (-2) = -1 \]

### Step 4: Solve for \( x \)
Simplifying the equation:
\[ -\frac{1}{\sqrt{x - 3}} = -1 \]
This gives:
\[ \frac{1}{\sqrt{x - 3}} = 1 \]
Taking the reciprocal:
\[ \sqrt{x - 3} = 1 \]
Squaring both sides:
\[ x - 3 = 1 \implies x = 4 \]

### Step 5: Find the corresponding \( y \) value
Now, substitute \( x = 4 \) back into the curve equation to find \( y \):
\[ y = \sqrt{4 - 3} = \sqrt{1} = 1 \]
Thus, we have one point \( (4, 1) \).

Since \( y = \sqrt{x - 3} \) can also be negative, we consider \( y = -1 \) as well. Therefore, the second point is \( (4, -1) \).

### Final Points
The points on the curve where the tangent is perpendicular to the line \( 6x + 3y - 5 = 0 \) are:
\[ (4, 1) \quad \text{and} \quad (4, -1) \]

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