The ellipse graphed below has equation
x^2 + 5y^2 = 8 − 2xy.
The tangent line pictured above has slope zero. Find the exact coordinates where it touches the ellipse.

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 \(y\).

The ellipse graphed below has equation x^2 + 5y^2 = 8 − 2xy. The tangent line pictured above has slope zero. Find the exact coordinates where it touches the ellipse.

Learn how to find the exact coordinates where a tangent line with slope zero touches the ellipse given by the equation x^2 + 5y^2 = 8 − 2xy. This detailed step-by-step solution uses implicit differentiation and covers key algebraic concepts. Suitable for advanced high school students in Grades 11-12.

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