Problem 34 - Calculus - AP Edition

Question

(a) A student claims that the ellipse \(x^{2}-x y+y^{2}=1\) has a horizontal tangent line at the point (1,1) Without doing any computations, explain why the student's claim must be incorrect. (b) Find all points on the ellipse \(x^{2}-x y+y^{2}=1\) at which the tangent line is horizontal.

Step-by-Step Solution

Answer

The claim is incorrect as the slope at (1,1) isn't zero. Points are \( \left( \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \right) \) and \( \left( -\frac{1}{\sqrt{3}}, -\frac{2}{\sqrt{3}} \right) \).

Step 1: Understanding the Claim

The student claims that the ellipse has a horizontal tangent line at the point (1,1). For a line to be horizontal, the slope of the tangent line must be zero. However, if a conic section has no vertical or smooth changes at a point, the slope cannot be zero. We check the given point to see if it fits the equation of the ellipse to start.

Step 2: Verification of the Point (1,1)

Substitute the point (1,1) into the ellipse equation to verify if it is on the curve: \[ 1^2 - 1 \cdot 1 + 1^2 = 1 \],which simplifies to \[ 1 - 1 + 1 = 1 \].This is true, so (1,1) is on the ellipse. However, this does not mean its tangent is horizontal.

Step 3: Horizontal Tangent Lines and Derivatives

To find if a tangent is horizontal, we set the derivative of the ellipse equation to zero. We'll implicitly differentiate the equation \( x^2 - xy + y^2 = 1 \) implicitly with respect to \( x \).

Step 4: Implicit Differentiation

Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(x^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1) \],leading to \[ 2x - (x \frac{dy}{dx} + y) + 2y \frac{dy}{dx} = 0 \]. Rearrange as \[ (2y - x) \frac{dy}{dx} = y - 2x \].

Step 5: Setting up the Horizontal Tangent Condition

For the tangent to be horizontal, \[ \frac{dy}{dx} = 0 \].Set \[ y - 2x = 0 \] which implies \[ y = 2x \].

Step 6: Substituting the Horizontal Condition into the Ellipse Equation

Substitute \( y = 2x \) into the ellipse equation \( x^2 - xy + y^2 = 1 \): Replacing \( y \) with \( 2x \): \[ x^2 - x(2x) + (2x)^2 = 1 \]simplifies to \[ x^2 - 2x^2 + 4x^2 = 1 \],or \[ 3x^2 = 1 \].

Step 7: Solving for x

We solve \[ 3x^2 = 1 \]which gives\[ x^2 = \frac{1}{3} \],so \[ x = \pm \frac{1}{\sqrt{3}} \].

Step 8: Finding Corresponding y-values

Using \( y = 2x \), find the corresponding \( y \)-values:For \( x = \frac{1}{\sqrt{3}} \), \( y = \frac{2}{\sqrt{3}} \).For \( x = -\frac{1}{\sqrt{3}} \), \( y = -\frac{2}{\sqrt{3}} \).

Step 9: Conclusion on Horizontal Tangent Points

Thus, the points on the ellipse where the tangent line is horizontal are:\( \left( \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \right) \) and \( \left( -\frac{1}{\sqrt{3}}, -\frac{2}{\sqrt{3}} \right) \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations that are not easily solved for one variable in terms of the other. In such cases, you need to differentiate the equation with respect to one of the variables while treating the other variable as an implicit function.
For example, consider the ellipse equation \( x^2 - xy + y^2 = 1 \). Here, \( y \) is treated as a function of \( x \), even though it is not explicitly solved as such. When you differentiate with respect to \( x \), apply the product rule to terms with mixed variables and chain rule to terms involving \( y \).

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( -xy \) involves the product rule, leading to \( -(x \frac{dy}{dx} + y) \).
• The derivative of \( y^2 \) involves the chain rule, leading to \( 2y \frac{dy}{dx} \).

This entire approach allows you to find \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \), in equations that involve both \( x \) and \( y \).

Ellipse Equation

An ellipse is a set of points where the sum of the distances from two fixed points, known as foci, is constant. The equation \( x^2 - xy + y^2 = 1 \) describes an ellipse, but in a slightly unconventional form.
Ellipses usually appear in the standard forms \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) or \( ax^2 + bxy + cy^2 = 1 \) with certain constraints. The presence of the \( xy \) term in our case shows a rotated or skewed ellipse, unlike the typical axis-aligned cases.
To analyze this ellipse and work with its constraints, it's vital to understand its geometry and how its equation relates to shapes you'll visualize. Recognizing how polynomial powers and coefficients affect the graph provides insights into the ellipse's orientation and proportions.

Derivative of Ellipse

Taking the derivative of an ellipse's equation is crucial for finding the tangent lines to the curve at any particular point.
When seeking horizontal tangents on the ellipse described by \( x^2 - xy + y^2 = 1 \), you need to solve for \( \frac{dy}{dx} = 0 \). This condition signifies that the change in \( y \) with respect to \( x \) is zero—indicative of a horizontal tangent.
Substitute \( y = 2x \) back into the original ellipse equation. This substitution stems from setting \( y - 2x = 0 \), derived from solving the differential equation for a horizontal tangent.

• Replace \( y \) with \( 2x \) to get \( x^2 - 2x^2 + 4x^2 = 1 \).
• This simplifies to \( 3x^2 = 1 \), leading to \( x^2 = \frac{1}{3} \).
• Thus, \( x = \pm \frac{1}{\sqrt{3}} \).

Finally, calculate corresponding \( y \)-values using \( y = 2x \), yielding two points with horizontal tangents. Understanding derivatives in this manner reveals critical analysis points on the curve, such as horizontal tangents that highlight the curve's behavior.