Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

The slope of the tangent line to the hyperbola $$ 2 x^{2}-7 y^{2}-35=0 $$ at two points on the hyperbola is \(-\frac{2}{3}\). What are the coordinates of the points of tangency?

Step-by-Step Solution

Answer

The points of tangency are (7, -3) and (-7, 3).

Step 1: Implicit Differentiation of the Hyperbola

Start by differentiating the given equation of the hyperbola implicitly with respect to \(x\). The equation is \(2x^2 - 7y^2 - 35 = 0\). Differentiating gives:\[\frac{d}{dx}(2x^2) - \frac{d}{dx}(7y^2) = 0\]This yields:\[4x - 14y \frac{dy}{dx} = 0\]Solve for \(\frac{dy}{dx}\) to find the slope of the tangent line in terms of \(x\) and \(y\). The slope \(\frac{dy}{dx}\) is given by:\[\frac{dy}{dx} = \frac{2x}{7y}\]

Step 2: Set the Slope Equal to Given Value

Now set the slope of the tangent line equal to the provided slope, which is \(-\frac{2}{3}\). Thus:\[\frac{2x}{7y} = -\frac{2}{3}\]Cross multiplying gives:\[6x = -14y\]Simplify to find the relationship between \(x\) and \(y\):\[3x = -7y\]\[y = -\frac{3}{7}x\]

Step 3: Substitute Relationship into Original Hyperbola Equation

Substitute \(y = -\frac{3}{7}x\) back into the original hyperbola equation \(2x^2 - 7y^2 - 35 = 0\). Substitute \(y\) to get:\[2x^2 - 7\left(-\frac{3}{7}x\right)^2 - 35 = 0\]This simplifies to:\[2x^2 - 7\left(\frac{9}{49}x^2\right) - 35 = 0\]\[2x^2 - \frac{63}{49}x^2 - 35 = 0\]

Step 4: Simplify and Solve for x

Simplify \(\frac{63}{49}\) to \(\frac{9}{7}\) and substitute back:\[2x^2 - \frac{9}{7}x^2 - 35 = 0\]Multiply everything by 7 to eliminate fractions:\[14x^2 - 9x^2 - 245 = 0\]Combine the like terms:\[5x^2 - 245 = 0\]Solve for \(x^2\):\[5x^2 = 245\]\[x^2 = 49\]Thus, \(x = \pm 7\).

Step 5: Find Corresponding y Values

Using the relationship \(y = -\frac{3}{7}x\), find corresponding \(y\) values for both \(x = 7\) and \(x = -7\):For \(x = 7\):\[y = -\frac{3}{7}(7) = -3\]For \(x = -7\):\[y = -\frac{3}{7}(-7) = 3\]

Step 6: List the Points of Tangency

Combine the values of \(x\) and \(y\) to form ordered pairs, representing the points of tangency on the hyperbola. The solutions are the coordinates \((7, -3)\) and \((-7, 3)\).

Key Concepts

Hyperbola Tangent

Understanding the concept of a tangent line in the context of a hyperbola is crucial. A tangent line is a straight line that touches a curve at a point or points, matching the curve's slope at that point. In the case of the hyperbola given by the equation \[ 2x^2 - 7y^2 - 35 = 0 \], its tangent lines will touch the hyperbola at certain distinct points without crossing it.
To find these tangent lines, we use implicit differentiation, which is a technique to differentiate equations that are not solved explicitly for one variable. It helps identify the slope of the tangent line in relation to both variables, \(x\) and \(y\).
For our hyperbola, differentiating gives:\[ 4x - 14y \frac{dy}{dx} = 0 \]
This simplifies to finding the derivative \( \frac{dy}{dx} \), revealing the slope of the tangent line as \( \frac{2x}{7y} \). This slope expression is central to identifying the points of tangency on the hyperbola. By setting this equal to the given slope \(-\frac{2}{3}\), we can find the relationship between \(x\) and \(y\) for the tangent points.

Slope of a Line

The slope of a line is a number that describes both the direction and the steepness of the line. In the context of tangents and curves, the slope tells you how the curve behaves at a very specific point. For our hyperbola problem, the tangent line has a specific slope of \(-\frac{2}{3}\). Understanding what this means for the hyperbola is essential.
Slope is calculated as the "rise over run" or the change in \(y\) over the change in \(x\). A negative slope, which we have here, indicates that as you move along the line from left to right, the line falls downwards.
To integrate this concept into the problem, once you differentiate implicitly to find the slope \( \frac{2x}{7y} \), this is set equal to \(-\frac{2}{3}\). Solving this equation, one can find a relationship \( y = -\frac{3}{7}x \) which helps in finding exact points where the hyperbola meets this tangent line.

• Given slope (\(-\frac{2}{3}\)) equated to derived slope expression
• Cross-multiplying can solve for values relating \(x\) and \(y\)

Ultimately, this relationship is pivotal to decipher where exactly these tangents touch the hyperbola, which brings us to the concept of points of tangency.

Points of Tangency

Points of tangency are precisely where a tangent line meets a curve. For a hyperbola such as \[2x^2 - 7y^2 - 35 = 0 \], finding these points involves substituting discovered relationships back into the original equation.
After determining that the relationship between \(x\) and \(y\) is \(y = -\frac{3}{7}x\), you substitute this relationship back into the equation of the hyperbola to solve for specific values of \(x\) and \(y\). This method yields the points of tangency.
First, simplify and solve:\[ 5x^2 = 245 \]\[ x^2 = 49 \]which gives \(x = \pm 7\). With these \(x\) values, you can then determine \(y\) by substituting back into the relationship, yielding \(y = -3\) for \(x = 7\) and \(y = 3\) for \(x = -7\).

• Solutions yield the points: \((7, -3)\) and \((-7, 3)\)
• These coordinates are where the slope condition is satisfied and the tangent touches the curve

Combining all this, the process intricately paints a clear image of how calculus and analytic geometry work together to solve such problems.