To solve this problem, we need to find the points on the curve $y^2 + x^2 = 3 - xy$ where the tangent lines are either horizontal or vertical. We can achieve this by using implicit differentiation.

Step 1: Implicit Differentiation

Given the equation: $y^2 + x^2 = 3 - xy,$ we differentiate both sides with respect to $x$.

Differentiating the left side: $\frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) = 2y \frac{dy}{dx} + 2x.$

Differentiating the right side: $\frac{d}{dx}(3 - xy) = -\frac{d}{dx}(xy) = -\left(y + x\frac{dy}{dx}\right).$

Putting it all together: $2y\frac{dy}{dx} + 2x = -(y + x\frac{dy}{dx}).$

Step 2: Solve for $\frac{dy}{dx}$

Combine like terms to solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx} + x\frac{dy}{dx} = -y - 2x,$ $(2y + x)\frac{dy}{dx} = -y - 2x,$ $\frac{dy}{dx} = \frac{-y - 2x}{2y + x}.$

Step 3: Identify Conditions for Horizontal and Vertical Tangents

• Horizontal tangent lines occur when $\frac{dy}{dx} = 0$, so we set the numerator to zero: $-y - 2x = 0 \implies y = -2x.$
• Vertical tangent lines occur when the denominator is zero: $2y + x = 0 \implies y = -\frac{x}{2}.$

Step 4: Find Points of Intersection

Substitute $y = -2x$ into the original equation: $(-2x)^2 + x^2 = 3 - x(-2x),$ $4x^2 + x^2 = 3 + 2x^2,$ $6x^2 = 3 \implies x^2 = \frac{1}{2} \implies x = \pm\frac{1}{\sqrt{2}}.$

For $x = \frac{1}{\sqrt{2}}$, $y = -2\left(\frac{1}{\sqrt{2}}\right) = -\sqrt{2}$. For $x = -\frac{1}{\sqrt{2}}$, $y = -2\left(-\frac{1}{\sqrt{2}}\right) = \sqrt{2}$.

Substituting $y = -\frac{x}{2}$: $\left(-\frac{x}{2}\right)^2 + x^2 = 3 - x\left(-\frac{x}{2}\right),$ $\frac{x^2}{4} + x^2 = 3 + \frac{x^2}{2},$ $\frac{5x^2}{4} = 3 \implies x^2 = \frac{12}{5} \implies x = \pm\sqrt{\frac{12}{5}}.$

For $x = \sqrt{\frac{12}{5}}, y = -\frac{\sqrt{\frac{12}{5}}}{2}$. For $x = -\sqrt{\frac{12}{5}}, y = \frac{\sqrt{\frac{12}{5}}}{2}$.

Step 5: Count the Points

There are 4 points:

1. $\left(\frac{1}{\sqrt{2}}, -\sqrt{2}\right)$
2. $\left(-\frac{1}{\sqrt{2}}, \sqrt{2}\right)$
3. $\left(\sqrt{\frac{12}{5}}, -\frac{\sqrt{\frac{12}{5}}}{2}\right)$
4. $\left(-\sqrt{\frac{12}{5}}, \frac{\sqrt{\frac{12}{5}}}{2}\right)$

So, there are 4 points with either horizontal or vertical tangent lines.

Would you like further details on any step?

Here are some related questions:

1. How do you find the tangent lines to implicit curves?
2. What is the significance of implicit differentiation?
3. How can you classify critical points using derivatives?
4. How do horizontal and vertical tangents relate to function behavior?
5. What are the applications of implicit differentiation in real-world scenarios?

Tip: Always check the conditions of the numerator and denominator separately when solving for horizontal and vertical tangents in implicit differentiation.