Chapter 10: Problem 53
Prove: The line tangent to the hyperbola $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$ at the point \(\left(x_{0}, y_{0}\right)\) has the equation $$ \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1 $$
Short Answer
Expert verified
The tangent line equation is derived from the point-slope form using the slope from implicit differentiation.
Step by step solution
01
Understand the Hyperbola Equation
The given hyperbola is represented by the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). This is a hyperbola centered at the origin with the x-axis as its transverse axis.
02
Identify the Point and its Relationship with the Hyperbola
The point \((x_0, y_0)\) lies on the hyperbola. This implies \(\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1\) holds.
03
Use Implicit Differentiation to Find the Slope
Differentiate the hyperbola equation implicitly with respect to x: \( \frac{d}{dx}\left(\frac{x^2}{a^2}\right) - \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = 0 \). Thus, \( \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \), solving for \( \frac{dy}{dx} \) gives the slope: \( \frac{dy}{dx} = \frac{b^2}{a^2} \frac{x}{y} \).
04
Determine the Slope at the Given Point
Substitute \(x = x_0\) and \(y = y_0\) into the expression for the slope to get the slope of the tangent line: \( m = \frac{b^2}{a^2} \frac{x_0}{y_0} \).
05
Equation of the Tangent Line in Point-Slope Form
The equation of a line in point-slope form is \( y - y_0 = m(x - x_0) \). Substituting \( m = \frac{b^2}{a^2} \frac{x_0}{y_0} \), we have \( y - y_0 = \frac{b^2}{a^2} \frac{x_0}{y_0} (x - x_0) \).
06
Rearrange to Match the Given Tangent Line Equation
Rearrange the point-slope form to express in standard form: \( y = \frac{b^2}{a^2} \frac{x_0}{y_0} x - \frac{b^2 x_0^2}{a^2 y_0} + y_0 \). Multiply through by \( y_0 \) and rearrange terms: \( \frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1 \), proving it matches the given tangent equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not expressed explicitly as \( y = f(x) \). In the context of our problem, this method is essential because the hyperbola is given as an implicit equation: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]This equation cannot be easily solved for \( y \) in terms of \( x \).
To differentiate implicitly, we take the derivative of each term with respect to \( x \), treating \( y \) as an implicit function that depends on \( x \). This means whenever we differentiate a term involving \( y \), we must apply the chain rule and multiply by \( \frac{dy}{dx} \).
For example, differentiating the left-hand side of the hyperbola equation, we use the product rule and chain rule:
To differentiate implicitly, we take the derivative of each term with respect to \( x \), treating \( y \) as an implicit function that depends on \( x \). This means whenever we differentiate a term involving \( y \), we must apply the chain rule and multiply by \( \frac{dy}{dx} \).
For example, differentiating the left-hand side of the hyperbola equation, we use the product rule and chain rule:
- \( \frac{d}{dx}\left(\frac{x^2}{a^2}\right) = \frac{2x}{a^2} \)
- \( \frac{d}{dx}\left(-\frac{y^2}{b^2}\right) = -\frac{2y}{b^2} \frac{dy}{dx} \)
Slope of Tangent Line
The slope of a tangent line to a curve at a specific point describes how steep the line is at that point. For the hyperbola, after performing implicit differentiation, we find the slope \( m \) by solving for \( \frac{dy}{dx} \).
The formula we obtain is:\[ \frac{dy}{dx} = \frac{b^2}{a^2} \frac{x}{y} \]
This expression depends on both \( x \) and \( y \). To find the slope at a particular point \( (x_0, y_0) \), we substitute these values into the equation:\[ m = \frac{b^2}{a^2} \frac{x_0}{y_0} \]
The formula we obtain is:\[ \frac{dy}{dx} = \frac{b^2}{a^2} \frac{x}{y} \]
This expression depends on both \( x \) and \( y \). To find the slope at a particular point \( (x_0, y_0) \), we substitute these values into the equation:\[ m = \frac{b^2}{a^2} \frac{x_0}{y_0} \]
- \( x_0 \) and \( y_0 \) are the coordinates of the point lying on the hyperbola.
- The slope \( m \) provides the steepness of the tangent line at \( (x_0, y_0) \).
Standard Form of Hyperbola
The standard form of a hyperbola centered at the origin with the x-axis as its transverse axis is an important concept. It helps us understand the geometric properties of the hyperbola. The form is:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Here:
The point \( (x_0, y_0) \) being on the hyperbola means it satisfies this standard equation, a critical part of finding the tangent line.
Here:
- \( a \) and \( b \) are the real numbers that determine the shape and orientation of the hyperbola.
- The equation describes a hyperbola with its transverse axis along the x-axis.
- The center is at the origin \( (0, 0) \).
The point \( (x_0, y_0) \) being on the hyperbola means it satisfies this standard equation, a critical part of finding the tangent line.
Tangent Line Equation
The tangent line equation at a point on a curve gives the precise linear approximation to the curve at that point. Starting with the slope, after determining \( m \), the equation of the tangent line in point-slope form is:\[ y - y_0 = m(x - x_0) \]
Substituting the slope \( m = \frac{b^2}{a^2} \frac{x_0}{y_0} \), we get:\[ y - y_0 = \frac{b^2}{a^2} \frac{x_0}{y_0} (x - x_0) \]
To match the form \( \frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1 \), we need to rearrange the equation. This involves simplifying and rearranging terms:
Substituting the slope \( m = \frac{b^2}{a^2} \frac{x_0}{y_0} \), we get:\[ y - y_0 = \frac{b^2}{a^2} \frac{x_0}{y_0} (x - x_0) \]
To match the form \( \frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1 \), we need to rearrange the equation. This involves simplifying and rearranging terms:
- Expand and simplify the initial equation.
- Multiply through by \( y_0 \) to eliminate the fraction.
- Rewriting all terms to align with the desired form.