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?

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).