Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

The curve \(x^{2}-x y+y^{2}=16\) is an ellipse centered at the origin and with the line \(y=x\) as its major axis. Find the equations of the tangent lines at the two points where the ellipse intersects the \(x\)-axis.

Step-by-Step Solution

Answer

The tangent lines are \(y = -x + 4\) and \(y = -x - 4\).

Step 1: Setting up the ellipse equation

The given ellipse equation is \(x^2 - xy + y^2 = 16\). This equation needs to be analyzed to locate the points where it intersects the \(x\)-axis (where \(y = 0\)).

Step 2: Solving for x-intercepts

Set \(y = 0\) in the equation \(x^2 - xy + y^2 = 16\), so it becomes \(x^2 = 16\). Solving for \(x\), we get \(x = \pm 4\). Hence, the ellipse intersects the \(x\)-axis at the points \((4, 0)\) and \((-4, 0)\).

Step 3: Finding the derivative

To find the tangent line, we need the derivative of the ellipse equation. Differentiate \(x^2 - xy + y^2 = 16\) implicitly: \(2x - (x \frac{dy}{dx} + y) + 2y \frac{dy}{dx} = 0\). Rearrange to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{x-y}{2y-x}\).

Step 4: Calculating slopes at intersection points

Substitute \((4, 0)\) and \((-4, 0)\) into the derivative \(\frac{dy}{dx} = \frac{x-y}{2y-x}\). For \((4, 0)\), \(\frac{dy}{dx} = \frac{4-0}{0-4} = -1\). For \((-4, 0)\), \(\frac{dy}{dx} = \frac{-4-0}{0+4} = -1\). The slope at both points is \(-1\).

Step 5: Writing the tangent line equations

Using the point-slope form of a line \(y - y_1 = m(x - x_1)\) with slope \(-1\) and y-intercept points \((4, 0)\) and \((-4, 0)\), we write the equations: \(y - 0 = -1(x - 4)\) simplifies to \(y = -x + 4\) and \(y - 0 = -1(x + 4)\) simplifies to \(y = -x - 4\).

Key Concepts

Ellipse

An ellipse is a curved shape that resembles a flattened circle. One key characteristic of an ellipse is its major and minor axes. The major axis is the longest diameter, while the minor axis is the shortest. In this exercise, the ellipse's equation is given as:

• \[ x^2 - xy + y^2 = 16 \]

This equation can be considered in the context of its geometric properties. This particular ellipse is centered at the origin \((0,0)\), making it symmetric about this point. The line \(y=x\) is the major axis, which means this line is the longest stretch across the ellipse.
The importance of the coordinate axes shows in the eloquence of how points of intersection create simplified results. The ellipse intersects the \(x\)-axis at specific points, which we need to understand the tangent lines' behavior at these intersections.

Tangent lines

Tangent lines are essential in calculus as they provide a snapshot of the slope of a curve at a specific point. A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. In other words, it has the same slope as the curve does at that particular location.

• A tangent line never cuts through the curve, it merely grazes it.
• Its equation can be derived using the point-slope formula once the slope is known.

In this exercise, we need to find the tangent lines of the ellipse at its intersections with the \(x\)-axis. At the intersections, the slopes of these tangent lines specifically help in understanding how the ellipse behaves near the axes. By calculating the slopes at these points and applying the point-slope formula:

• For the point \((4, 0)\), the tangent line equation is \(y = -x + 4\).
• For the point \((-4, 0)\), the tangent line equation is \(y = -x - 4\).

Implicit differentiation

Implicit differentiation is a technique used in calculus to find the derivative of functions that are not easy to separate into explicit formulations (i.e., \(y\) cannot be easily isolated). With implicit differentiation, you can take derivatives of both sides of an equation with respect to \(x\) and then solve for \(\frac{dy}{dx}\).

• This technique helps in understanding rates of change even when the relationship isn't straightforward.
• It's a versatile method making it easier to differentiate complicated equations like ellipses, circles, and other lines.

For this ellipse equation, \(x^2 - xy + y^2 = 16\), the implicit differentiation process involves differentiating each term separately and then rearranging terms to solve for \(\frac{dy}{dx}\). By rewriting the relationship between \(x\) and \(y\), we ensure the slope of any tangent line can be accurately determined at any specified point.

X-intercepts

X-intercepts are the points where a curve crosses the x-axis. At these points, the \(y\)-value is always zero. Finding x-intercepts involves setting \(y=0\) in your equation and solving for \(x\).
In our ellipse example, we do this by plugging \(y=0\) into the equation:

• For \(x^2 - xy + y^2 = 16\), with \(y=0\), we simplify it to \(x^2 = 16\).
• Solving gives \(x = \pm 4\).

Therefore, the ellipse intersects the \(x\)-axis at the points \((4,0)\) and \((-4,0)\). These intercepts are crucial, as they serve as the locations where tangent lines will form, giving insight into the geometry and calculus of the ellipse as it interacts with the axes.