The equation of the tangent at (- 4, - 4) on the curve `x^2=4y` is

Options

A. 2x+y=4
B. 2x-y=12
C. 2x+y=-4
D. 2x-y=-4

Answer

To find the equation of the tangent at the point (-4, -4) on the curve given by \( x^2 = 4y \), we will follow these steps:

### Step 1: Find the derivative of the curve
The curve is given by the equation \( x^2 = 4y \). We need to differentiate this equation with respect to \( x \).

\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4y) \]

This gives us:

\[ 2x = 4\frac{dy}{dx} \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation from Step 1, we can isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} \]

### Step 3: Evaluate the derivative at the point (-4, -4)
Now we will substitute \( x = -4 \) into the derivative to find the slope \( m \) of the tangent line at the point (-4, -4):

\[ m = \frac{-4}{2} = -2 \]

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

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

Substituting \( m = -2 \), \( x_1 = -4 \), and \( y_1 = -4 \):

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

This simplifies to:

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

### Step 5: Simplify the equation
Now we will simplify the equation:

\[ y + 4 = -2x - 8 \]

Subtracting 4 from both sides gives:

\[ y = -2x - 12 \]

### Step 6: Rearranging to standard form
To express the equation in standard form, we can rearrange it:

\[ 2x + y + 12 = 0 \]

### Final Answer
The equation of the tangent at the point (-4, -4) on the curve \( x^2 = 4y \) is:

\[ 2x + y + 12 = 0 \]