To find the exact coordinates where the tangent line touches the ellipse given the equation:

$x^2 + 5y^2 = 8 - 2xy$

and the fact that the tangent line has a slope of zero, we begin by interpreting what the slope of zero means: the tangent line is horizontal, implying that the point where it touches the ellipse will have a slope of 0 (i.e., $\frac{dy}{dx} = 0$).

Step 1: Implicit differentiation

We'll first differentiate the given equation implicitly with respect to $x$:

$x^2 + 5y^2 = 8 - 2xy$

Differentiate both sides:

1. The derivative of $x^2$ with respect to $x$ is $2x$.
2. The derivative of $5y^2$ with respect to $x$ is $10y \frac{dy}{dx}$ (using the chain rule).
3. The derivative of the right side $8 - 2xy$ with respect to $x$ is $-2 \left( y + x \frac{dy}{dx} \right)$ (using the product rule for $xy$).

So we get:

$2x + 10y \frac{dy}{dx} = -2 \left( y + x \frac{dy}{dx} \right)$

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

Rearrange the equation to group all terms with $\frac{dy}{dx}$ on one side:

$10y \frac{dy}{dx} + 2x \frac{dy}{dx} = -2y - 2x$

Factor out $\frac{dy}{dx}$:

$(10y + 2x) \frac{dy}{dx} = -2y - 2x$

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

$\frac{dy}{dx} = \frac{-2(y + x)}{10y + 2x}$

Step 3: Set $\frac{dy}{dx} = 0$

Since the tangent line has a slope of 0, set $\frac{dy}{dx} = 0$:

$\frac{-2(y + x)}{10y + 2x} = 0$

For the fraction to equal 0, the numerator must be 0:

$y + x = 0$

Thus, $y = -x$.

Step 4: Substitute $y = -x$ into the original equation

Now substitute $y = -x$ into the original equation $x^2 + 5y^2 = 8 - 2xy$:

$x^2 + 5(-x)^2 = 8 - 2x(-x)$

Simplify:

$x^2 + 5x^2 = 8 + 2x^2$

$6x^2 = 8 + 2x^2$

Subtract $2x^2$ from both sides:

$4x^2 = 8$

Solve for $x^2$:

$x^2 = 2$

Thus, $x = \pm \sqrt{2}$.

Step 5: Find the corresponding $y$ values

Since $y = -x$, the corresponding $y$ values are:

$y = -\sqrt{2} \quad \text{or} \quad y = \sqrt{2}.$

Conclusion

The points where the tangent line touches the ellipse are:

$(\sqrt{2}, -\sqrt{2}) \quad \text{and} \quad (-\sqrt{2}, \sqrt{2}).$

Would you like more details on any step?

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Here are 5 related questions to deepen your understanding:

1. How do you find the derivative of an implicit function in general?
2. What does a horizontal tangent line indicate about a function’s derivative?
3. Can you find the points of intersection of an ellipse and a line?
4. How would the process change if the tangent line had a non-zero slope?
5. How can you classify an ellipse and its properties from its equation?

Tip: When dealing with implicit differentiation, always apply the product rule carefully when differentiating terms involving products of $x$ and $y$.