Question

if $-xy - 4 - y = 0$ then find the equations of all tangent lines to the curve when $x = -3$.

Explanation:

Step1: Find the value of y when x = -3

Substitute \( x = -3 \) into the equation \( -xy - 4 - y = 0 \):
\( -(-3)y - 4 - y = 0 \)
Simplify: \( 3y - 4 - y = 0 \)
Combine like terms: \( 2y - 4 = 0 \)
Add 4 to both sides: \( 2y = 4 \)
Divide by 2: \( y = 2 \)

Step2: Implicit differentiation to find \( \frac{dy}{dx} \)

Start with \( -xy - 4 - y = 0 \)
Differentiate both sides with respect to x:
Using product rule on \( -xy \): \( - (y + x\frac{dy}{dx}) - 0 - \frac{dy}{dx} = 0 \)
Simplify: \( -y - x\frac{dy}{dx} - \frac{dy}{dx} = 0 \)
Collect terms with \( \frac{dy}{dx} \): \( -x\frac{dy}{dx} - \frac{dy}{dx} = y \)
Factor out \( \frac{dy}{dx} \): \( \frac{dy}{dx}(-x - 1) = y \)
Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{y}{-x - 1} = \frac{-y}{x + 1} \)

Step3: Find the slope at x = -3, y = 2

Substitute \( x = -3 \) and \( y = 2 \) into \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{-2}{-3 + 1} = \frac{-2}{-2} = 1 \)

Step4: Use point - slope form to find the tangent line equation

Point - slope form is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(-3,2) \) and \( m = 1 \)
\( y - 2 = 1\times(x + 3) \)
Simplify: \( y - 2 = x + 3 \)
\( y = x + 5 \)

Answer:

The equation of the tangent line is \( y = x + 5 \)