Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equations of the tangent lines to the ellipse \(x^{2}+2 y^{2}-2=0\) that are parallel to the line $$ 3 x-3 \sqrt{2} y-7=0 $$

Step-by-Step Solution

Answer

The tangent line equations are: 1. \(y - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x + 1)\) 2. \(y + \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x - 1)\)

Step 1: Identify the slope of the given line

The equation of the line is given as \(3x - 3\sqrt{2}y - 7 = 0\). Rewrite it in slope-intercept form (\(y = mx + c\)) to find the slope \(m\). This gives: \(y = \frac{3}{3\sqrt{2}}x - \frac{7}{3\sqrt{2}}\). Thus, the slope of the given line is \(m = \frac{\sqrt{2}}{2}\).

Step 2: Determine the general tangent line equation

To find the equation of the tangent line to the ellipse \(x^2 + 2y^2 = 2\), use the implicit differentiation technique. The derivative \(\frac{dy}{dx}\) will allow us to form the tangent line equation.

Step 3: Calculate the derivative of the ellipse

Implicitly differentiate the equation \(x^2 + 2y^2 = 2\) with respect to \(x\). This gives: \(2x + 4yy' = 0\). Solving for \(y'\) (the slope of the tangent line), we get \(y' = -\frac{x}{2y}\).

Step 4: Set the tangent line slope equal to the given slope

To find the points of tangency, equate the derivative \(y' = -\frac{x}{2y}\) to the slope of the given line \(\frac{\sqrt{2}}{2}\). Solve for \(x\) and \(y\) in the equation \(-\frac{x}{2y} = \frac{\sqrt{2}}{2}\), resulting in \(x = -\sqrt{2}y\).

Step 5: Substitute back into the ellipse equation

Substitute \(x = -\sqrt{2}y\) back into the ellipse equation \(x^2 + 2y^2 = 2\). This gives \((\sqrt{2}y)^2 + 2y^2 = 2\), which simplifies to \(4y^2 = 2\). Solving for \(y\) gives \(y = \pm\frac{1}{\sqrt{2}}\).

Step 6: Calculate corresponding x-values

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

Step 7: Form the tangent equations

The points of tangency are \((-1, \frac{1}{\sqrt{2}})\) and \((1, -\frac{1}{\sqrt{2}})\). Use the point-slope form \(y - y_1 = m(x - x_1)\) with \(m = \frac{\sqrt{2}}{2}\) to find the tangent equations. The tangents are: 1. \(y - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x + 1)\)2. \(y + \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x - 1)\)

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of equations not solved for one variable in terms of another. In this case, we deal with the ellipse equation \(x^2 + 2y^2 = 2\). Instead of isolating \(y\) and differentiating, we directly differentiate both sides with respect to \(x\). This requires the use of the chain rule, because \(y\) is also a function of \(x\).
When you derive the left side, you get \(2x + 4yy' = 0\). Notice that \(y'\), called the derivative of \(y\) with respect to \(x\), appears because \(y\) changes with \(x\). By solving for \(y'\), we find the slope of the tangent line to the ellipse at any given point. In this problem, that slope is \(y' = -\frac{x}{2y}\). This equation is crucial because it gives us a formula for the slope that is parallel to our target line.

Slope of a Line

The slope of a line is essentially a measure of how steep the line is. In the context of the given exercise, you need to find the slope of the line \(3x - 3\sqrt{2}y - 7 = 0\) first. To do this, we rearrange it to the form \(y = mx + c\), where \(m\) is the slope.
To convert the line equation to slope-intercept form, isolate \(y\): \(y = \frac{3}{3\sqrt{2}}x - \frac{7}{3\sqrt{2}}\). Here, the slope \(m\) is \(\frac{\sqrt{2}}{2}\). Since two lines are parallel if they have the same slope, determining this value was essential to find lines tangent to the ellipse that are parallel to the given line.

Point-Slope Form Equation

The point-slope form of a line equation is handy when you know a line's slope and a point through which it passes. This form is represented as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point and \(m\) is the slope. It directly derives the line equation from these known values.
In the exercise, once we identified the slope \(m = \frac{\sqrt{2}}{2}\) and the points of tangency \((-1, \frac{1}{\sqrt{2}})\) and \((1, -\frac{1}{\sqrt{2}})\), the point-slope form was used to derive the equations of the lines tangent to the ellipse:

• \(y - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x + 1)\)
• \(y + \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}(x - 1)\)

These are the equations of the desired tangent lines parallel to the given line.

Parallel Lines

Parallel lines are lines in a plane that, although extended in both directions indefinitely, never intersect. For two lines to be parallel, they must have the same slope. This was a goal in the given problem - to find tangent lines to the ellipse that are parallel to a specific line.
We've already determined the slope of the required parallel line \( \left(m = \frac{\sqrt{2}}{2}\right) \). We then found the derivative of the ellipse and set its slope equal to \( \frac{\sqrt{2}}{2} \). By solving this, we managed to discover the points of tangency. Hence, using the point-slope formula, we found lines that not only satisfy the tangency conditions but also remain parallel to the provided line.