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We are given the equation:

\[
-xy - y^2 + 4 = 2x^2
\]

and the derivative of \( y \) with respect to \( x \):

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

We are tasked with finding all points where \( x = -2 \) and the tangent to the curve is vertical.

### Step 1: Vertical Tangent Condition
A vertical tangent occurs when \( \frac{dx}{dy} = 0 \), which is equivalent to the denominator of the derivative \( \frac{dy}{dx} \) being zero. From the given formula for \( \frac{dy}{dx} \), we have:

\[
x - 2y = 0
\]

Thus, for a vertical tangent, we have:

\[
x = 2y
\]

### Step 2: Substituting \( x = -2 \) and \( x = 2y \)
We are specifically interested in the case where \( x = -2 \), so let's substitute \( x = -2 \) into the condition \( x = 2y \):

\[
-2 = 2y \quad \Rightarrow \quad y = -1
\]

Thus, we have the point \( (-2, -1) \).

### Step 3: Verify the Point Satisfies the Original Equation
Now, we need to verify whether the point \( (-2, -1) \) satisfies the original equation \( -xy - y^2 + 4 = 2x^2 \). Substituting \( x = -2 \) and \( y = -1 \) into the equation:

\[
-(-2)(-1) - (-1)^2 + 4 = 2(-2)^2
\]

Simplifying:

\[
-2 - 1 + 4 = 8 \quad \Rightarrow \quad 1 = 8
\]

This is false, meaning that the point \( (-2, -1) \) does not satisfy the original equation.

### Step 4: Conclusion
Since \( (-2, -1) \) does not satisfy the original equation, there are no points on the curve where \( x = -2 \) and the tangent is vertical.

### Summary
There are no such points on the curve where \( x = -2 \) and the tangent to the curve is vertical.

Would you like further details on any specific step?

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Here are some follow-up questions:

1. How do we determine if a tangent line is vertical in general?
2. Can you explain the geometric significance of a vertical tangent line?
3. What are some other ways to find critical points on a curve?
4. How do we derive \( \frac{dy}{dx} \) from implicit differentiation for such an equation?
5. What are the conditions for a curve to have a horizontal tangent?

**Tip:** When dealing with implicit differentiation, always solve for \( \frac{dy}{dx} \) before checking specific conditions (e.g., vertical tangents).

If -xy - y^2 + 4 = 2x^2 and it is known that dy/dx = (y + 4x)/(x - 2y), find all coordinate points on the curve where x = -2 and the line tangent to the curve is vertical, or state that no such points exist.

Solve for all coordinate points where x = -2 and the tangent to the curve -xy - y^2 + 4 = 2x^2 is vertical using implicit differentiation. Understand how to set conditions for vertical tangents and verify points. Suitable for students in Grades 11-12.

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