Question

(a) Use implicit differentiation to find an equation of the tangent line to the hyperbola \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\) at \((3,-2)\). (b) Show that the equation of the tangent line to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is \(\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1\).

Short answer

The equation of the tangent line to the hyperbola \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\) at \(3,-2\) is \(y=-2-\frac{x-3}{4}\). The equation of the tangent line to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is \(y=\frac{x_0x}{a^{2}}-\frac{y_0^{2}}{b^{2}}+\frac{x_0^{2}}{a^{2}}\).

Step-by-step solution

Differentiate the hyperbola equation

Apply the power rule of differentiation which is \((x^n)' = nx^{n-1}\) to each term. The derivative of \(\frac{x^{2}}{6}\) is \(\frac{1}{3}x\) and the derivative of \(y^{2}\) is \(2yy'\). Substituting these you get: \(\frac{1}{3}x - \frac{2yy'}{8} = 0\).

Solve for y'

Rearrange the equation from step 1 to find y': \(y' = \frac{1}{6}x/y\).

Find the slope of the tangent line

Substitute the point (3,-2) into y' to find the slope (m) of the tangent line: \(m=y'(3,-2)=\frac{1}{6}*3/-2=-\frac{1}{4}\). The equation of the line will be in the form \(y-y_1=m(x-x_1)\) where \((x_1,y_1)=(3,-2)\). Substituting these values will yield the equation \(y=-2-\frac{1}{4}(x-3)\).

Differentiate the general hyperbola equation

The derivative of \(\frac{x^{2}}{a^{2}}\) is \(\frac{2x}{a^{2}}\) and the derivative of \(\frac{y^{2}}{b^{2}}\) is \(\frac{2yy'}{b^{2}}\). We then have \(\frac{2x}{a^{2}} - \frac{2yy'}{b^{2}} = 0\). Solving for y' gives \(y' = \frac{x}{a^{2}y}\).

Find the general equation of the tangent line

The general equation of the tangent line is given by \(y-y_0=y'(x_0,y_0)(x-x_0)\). With \(y' = \frac{x_0}{a^{2}y_0}\), the equation of the tangent becomes \(y-y_0=\frac{x_0}{a^{2}y_0}(x-x_0)\). Solving for y we get the equation of the tangent line to the hyperbola at \((x_0, y_0)\) is \(y=\frac{x_0x}{a^{2}}-\frac{y_0^{2}}{b^{2}}+\frac{x_0^{2}}{a^{2}}\).

Key concepts

Tangent Line Equation

Understanding the equation of a tangent line to a curve is crucial in calculus, as it provides insight into the function's behavior at a specific point. A tangent line to a curve at a given point is a straight line that touches the curve exactly at that point and has the same slope as the curve at that point.

To find the equation of a tangent line, we generally use the point-slope form of a line, which is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the tangent line, and \(x_1, y_1\) are the coordinates of the point of tangency. To solve for this equation using implicit differentiation, which is especially handy when dealing with curves not easily solvable for \(y\), differentiate both sides of the equation with respect to \(x\), rearrange to solve for \(y'\) (the derivative of \(y\)), and then plug in the coordinates of the point in question to find the slope \(m\) at that point.

Once the slope is known, and you have the coordinates of the point of tangency, the tangent line equation can be written directly. For the hyperbola example in the exercise, the point \( (3, -2) \) and slope \(m = -\frac{1}{4}\) allowed us to define the tangent line equation. This tangible method lays out a clear path to follow for students seeking to understand the steps involved in deriving the tangent line equation for any point on a curve.

Hyperbola Differentiation

A hyperbola, a type of conic section, can be represented by its general equation \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). When we differentiate a hyperbola implicitly with respect to \(x\), we apply differentiation rules to each term involving \(x\) or \(y\).

In such cases, terms containing \(y\) are differentiated using the chain rule, because \(y\) is a function of \(x\). This means that for the term \(\frac{y^{2}}{b^{2}}\), its differentiation involves \(y'\), the derivative of \(y\) with respect to \(x\), resulting in \(\frac{2yy'}{b^{2}}\). By setting the derivative equal to zero and solving for \(y'\), we find the slope of the tangent line to the hyperbola at any point \( (x_0, y_0) \).

This approach to differentiation is essential for accurately finding slopes of tangent lines to hyperbolas and other curves where direct differentiation isn't possible. When students apply these steps correctly, they increase their proficiency in dealing with implicit differentiation and understanding the slopes of tangent lines on hyperbolas at various points.

Slope of Tangent

The slope of the tangent to a curve at a given point provides a numerical measure of the curve's steepness at that precise location. It is the derivative of a function at the point of tangency, denoted as \(y'\) or \(\frac{dy}{dx}\).

Finding the slope of a tangent is a fundamental skill in calculus, which is initially taught by using explicit functions. However, with implicit differentiation, we can find the slope even when the function is not explicitly solved for \(y\). After differentiating the equation of the curve with respect to \(x\), the resulting expression allows us to solve for \(y'\) and substitute the coordinates of the point of interest. This calculation yields the specific slope of the tangent at that point. In the case of the hyperbola, the slope at point \( (3, -2) \) was found to be \(m = -\frac{1}{4}\), which indicates that the tangent to the curve at that point is descending to the right.

By mastering the calculation of the slope of the tangent, students can predict and visualize the behavior of various functions at specific points. Effective understanding of this concept aids in grasping other aspects of calculus, such as optimization and curve sketching. Hence, emphasizing its calculation in different contexts is valuable for a student’s mathematical development.