Find the equations of the tangent and the normal to the curve `y^2=(x^3)/(4-x)` at `(2,\ -2)` at the indicated points

Explanation

To find the equations of the tangent and the normal to the curve \( y^2 = \frac{x^3}{4 - x} \) at the point \( (2, -2) \), we will follow these steps:

### Step 1: Differentiate the curve
We start by differentiating the given equation with respect to \( x \):
\[ y^2 = \frac{x^3}{4 - x} \]
Using implicit differentiation, we differentiate both sides:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}\left(\frac{x^3}{4 - x}\right) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = \frac{(4 - x)(3x^2) - x^3(-1)}{(4 - x)^2} \]
This simplifies to:
\[ 2y \frac{dy}{dx} = \frac{(4 - x)(3x^2) + x^3}{(4 - x)^2} \]

### Step 2: Substitute the point into the derivative
Next, we substitute the point \( (2, -2) \) into the derivative to find the slope of the tangent line:
\[ 2(-2) \frac{dy}{dx} = \frac{(4 - 2)(3(2^2)) + 2^3}{(4 - 2)^2} \]
Calculating the right side:
\[ 2(-2) \frac{dy}{dx} = \frac{(2)(12) + 8}{(2)^2} = \frac{24 + 8}{4} = \frac{32}{4} = 8 \]
Thus, we have:
\[ -4 \frac{dy}{dx} = 8 \implies \frac{dy}{dx} = -2 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-2} = \frac{1}{2} \]

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
For the tangent line:
\[ y - (-2) = -2(x - 2) \]
This simplifies to:
\[ y + 2 = -2x + 4 \implies y = -2x + 2 \]

### Step 5: Write the equation of the normal line
Using the point-slope form for the normal line:
\[ y - (-2) = \frac{1}{2}(x - 2) \]
This simplifies to:
\[ y + 2 = \frac{1}{2}x - 1 \implies y = \frac{1}{2}x - 3 \]

### Final Equations
Thus, the equations of the tangent and normal lines at the point \( (2, -2) \) are:
- Tangent: \( y = -2x + 2 \)
- Normal: \( y = \frac{1}{2}x - 3 \)