Ch.12 - Parametric and Polar Curves

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch.12 - Parametric and Polar Curves

Problem 12.4.74

Chapter 12, Problem 12.4.74

Tangent lines for a hyperbola Find an equation of the line tangent to the hyperbola x²/a² + y²/b² = 1 at the point (x₀, y₀)

Verified step by step guidance

Start with the given hyperbola equation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Implicitly differentiate both sides of the equation with respect to \(x\) to find \(\frac{dy}{dx}\). Use the chain rule for the \(y^2\) term, treating \(y\) as a function of \(x\).

After differentiating, you will get: \(\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0\). Solve this equation for \(\frac{dy}{dx}\) to find the slope of the tangent line at any point \((x, y)\) on the hyperbola.

Substitute the given point \((x_0, y_0)\) into the expression for \(\frac{dy}{dx}\) to find the slope of the tangent line at that specific point.

Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope found in the previous step, to write the equation of the tangent line at \((x_0, y_0)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is used to find the derivative of y with respect to x when y is defined implicitly by an equation involving both variables. For the hyperbola equation, it allows us to differentiate both sides with respect to x to find the slope of the tangent line at a given point.

Equation of a Tangent Line

The tangent line to a curve at a point has a slope equal to the derivative of the curve at that point. Once the slope is found, the tangent line equation can be written using the point-slope form: y - y₀ = m(x - x₀), where m is the slope and (x₀, y₀) is the point of tangency.

Hyperbola Equation and Properties

A hyperbola is defined by the equation x²/a² - y²/b² = 1 (or similar forms). Understanding its standard form and properties helps identify the correct implicit differentiation steps and interpret the geometric meaning of the tangent line at a specific point on the curve.

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