For the curve `x^(2)+ 4xy + 8y^(2)`= 64` the tangents are parallel to the x-axis only at the points

Options

A. `(0,2sqrt(2)` and `(0,-2sqrt(2))`
B. (8,-4) and (-8,4)
C. `(8sqrt(2),-2sqrt(2))` and `(-8sqrt(2),2sqrt(2))`
D. (8,0) and (-8,0)

Answer

To find the points on the curve \( x^2 + 4xy + 8y^2 = 64 \) where the tangents are parallel to the x-axis, we need to follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ x^2 + 4xy + 8y^2 = 64 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(4xy) + \frac{d}{dx}(8y^2) = \frac{d}{dx}(64) \]
This gives us:
\[ 2x + 4\left(y + x\frac{dy}{dx}\right) + 16y\frac{dy}{dx} = 0 \]

### Step 2: Simplify the differentiation
Expanding the differentiation:
\[ 2x + 4y + 4x\frac{dy}{dx} + 16y\frac{dy}{dx} = 0 \]
Now, we can group the terms involving \( \frac{dy}{dx} \):
\[ 2x + 4y + (4x + 16y)\frac{dy}{dx} = 0 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging the equation to isolate \( \frac{dy}{dx} \):
\[ (4x + 16y)\frac{dy}{dx} = - (2x + 4y) \]
Thus,
\[ \frac{dy}{dx} = -\frac{2x + 4y}{4x + 16y} \]

### Step 4: Set the derivative to zero for horizontal tangents
For the tangent to be parallel to the x-axis, we set \( \frac{dy}{dx} = 0 \):
\[ -\frac{2x + 4y}{4x + 16y} = 0 \]
This implies:
\[ 2x + 4y = 0 \]
From this, we can simplify:
\[ x + 2y = 0 \quad \Rightarrow \quad x = -2y \]

### Step 5: Substitute \( x = -2y \) into the original equation
Now we substitute \( x = -2y \) back into the original curve equation:
\[ (-2y)^2 + 4(-2y)y + 8y^2 = 64 \]
This simplifies to:
\[ 4y^2 - 8y^2 + 8y^2 = 64 \]
Combining like terms:
\[ 4y^2 = 64 \]
Dividing both sides by 4:
\[ y^2 = 16 \quad \Rightarrow \quad y = 4 \quad \text{or} \quad y = -4 \]

### Step 6: Find corresponding \( x \) values
Using \( y = 4 \):
\[ x = -2(4) = -8 \]
Using \( y = -4 \):
\[ x = -2(-4) = 8 \]

### Final Points
The points where the tangents to the curve are parallel to the x-axis are:
\[ (-8, 4) \quad \text{and} \quad (8, -4) \]