Find the equation of the tangent line to the graph of equation √x + √(y=3) at the point (1,4).

Question 14 options:

1) 

None of these.


2) 

y - 4 = 2(x - 1)


3) 

y - 1 = -2(x - 4)


4) 

y-4 = (-1/2)(x-1)

The given equation is:

\[
\sqrt{x} + \sqrt{y} = 3
\]

We are asked to find the equation of the tangent line to this curve at the point \((1, 4)\).

### Step 1: Differentiate implicitly
We will differentiate both sides of the equation with respect to \(x\).

First, rewrite the equation for easier differentiation:

\[
\sqrt{x} + \sqrt{y} = 3
\]

The derivative of \(\sqrt{x}\) is:

\[
\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}
\]

The derivative of \(\sqrt{y}\), using the chain rule, is:

\[
\frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx}
\]

Now, differentiate both sides:

\[
\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearrange the equation to isolate \(\frac{dy}{dx}\):

\[
\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}
\]

Multiply both sides by \(2\sqrt{y}\):

\[
\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}
\]

### Step 3: Evaluate the derivative at \((1, 4)\)
Substitute \(x = 1\) and \(y = 4\) into the derivative:

\[
\frac{dy}{dx} = -\frac{\sqrt{4}}{\sqrt{1}} = -\frac{2}{1} = -2
\]

Thus, the slope of the tangent line at the point \((1, 4)\) is \(-2\).

### Step 4: Use point-slope form to find the equation of the tangent line
The point-slope form of a line is given by:

\[
y - y_1 = m(x - x_1)
\]

Where:
- \(m\) is the slope (\(-2\)),
- \((x_1, y_1)\) is the point of tangency \((1, 4)\).

Substitute these values into the formula:

\[
y - 4 = -2(x - 1)
\]

This simplifies to:

\[
y - 4 = -2(x - 1)
\]

### Step 5: Compare with the given options
The correct answer is:

\[
\boxed{y - 4 = -2(x - 1)}
\]

However, this option is not directly listed, so the closest option, assuming there is an error in the list, is:

Option **3: None of these.**

---

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you apply implicit differentiation to more complex equations?
2. What is the geometric interpretation of a tangent line?
3. How does the chain rule apply in implicit differentiation?
4. How would the process change if we were finding a normal line instead of a tangent line?
5. Can we always find a tangent line for any point on a curve?

**Tip:** Implicit differentiation is especially useful when you cannot easily solve for one variable in terms of the other.

Find the equation of the tangent line to the graph of equation √x + √(y=3) at the point (1,4).

Learn how to find the equation of the tangent line to the graph of √x + √y = 3 at the point (1, 4) using implicit differentiation. This problem covers key concepts such as the chain rule and differentiation of square roots, suitable for Grades 11-12.

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