Question

Find an equation of the line tangent to the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\)

Short answer

Question: Find the equation of the tangent line to the hyperbola \(x^2/a^2 - y^2/b^2 = 1\) at the point \((x_0, y_0)\). Answer: The equation of the tangent line to the given hyperbola at the point \((x_0, y_0)\) is \(y = \frac{x_0}{b^2}(x - x_0) + y_0\).

Step-by-step solution

Implicitly differentiate the equation with respect to x

First, we want to differentiate the equation \(x^2/a^2 - y^2/b^2 = 1\) with respect to \(x\). Using implicit differentiation, we have: \(\frac{d}{dx}\left(\frac{x^2}{a^2}\right) - \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = \frac{d}{dx}\left(1\right)\) Using the chain rule, we obtain: \(\frac{2x}{a^2} - \frac{2yy'}{b^2} = 0\)

Solve for y'

Now, we want to isolate \(y'\) in our expression to find the derivative. Solving for \(y'\), we have: \(y' = \frac{xy}{a^2b^2}\left(\frac{a^2}{y}\right)\) Now, simplifying the expression, we find: \(y' = \frac{x}{b^2}\)

Find the slope at the point (x_0, y_0)

Now, we substitute the point \((x_0, y_0)\) into our expression for \(y'\) to find the slope of the tangent line at that point: \(m = y'(x_0) = \frac{x_0}{b^2}\)

Use the point-slope form to find the equation of the tangent line

The point-slope form of a linear equation is given by: \(y - y_0 = m(x - x_0)\) Substituting the values we found for \(m, x_0,\) and \(y_0\), we have: \(y - y_0 = \frac{x_0}{b^2}(x - x_0)\) Now, to find the equation of the tangent line, we simply rearrange the terms: \(y = \frac{x_0}{b^2}(x - x_0) + y_0\)

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique used in calculus to find derivatives of functions that are not expressed explicitly. In cases where we have an equation involving both variables, like the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), implicit differentiation offers a way to differentiate without solving for one variable in terms of the other.
The key steps involve:

• Applying the differentiation operator \(\frac{d}{dx}\) to both sides of the equation.
• Using the chain rule, which allows us to differentiate composite functions. For example, differentiating \(\frac{y^2}{b^2}\) with respect to \(x\) involves the expression \(-\frac{2y}{b^2}y'\), acknowledging that \(y\) is a function of \(x\).
• Solving for \(y'\), which represents the derivative of \(y\) with respect to \(x\).

By isolating \(y'\), you can determine the slope of a curve at any point \((x_0, y_0)\). Implicit differentiation is especially useful for finding tangent slopes that aren't immediately obvious from explicit functions.

Hyperbola

A hyperbola is a type of conic section formed by the intersection of a plane with both nappes of a double cone. Its standard equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) describes a hyperbola centered at the origin with horizontal transverse axes.
Some key points about hyperbolas include:

• They consist of two symmetrical curves called branches.
• The values \(a\) and \(b\) determine the distances from the center to the vertices and foci, respectively.

When finding a tangent to a hyperbola, understanding its geometric properties is essential. This allows us to accurately apply differentiation techniques at a given point \((x_0, y_0)\). Moreover, hyperbolas exhibit interesting properties such as having asymptotes with equations \(y = \pm \frac{b}{a}x\), which guide the directions of its branches.

Point-Slope Form

The point-slope form of a linear equation is extremely helpful when you need to write the equation of a line given a point and a slope. It is expressed as:
\[ y - y_0 = m(x - x_0) \]
This form simplifies finding the equation of a tangent line after determining the slope \(m\) from implicit differentiation. Here are the steps:

• Identify the slope \(m\) at a specific point \((x_0, y_0)\), like in a calculus problem.
• Use the known point and the slope to plug into the point-slope formula.
• Solve for \(y\) to get the equation in a usable format.

In the case of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we find the slope using implicit differentiation and substitute it into our point-slope equation. This method provides a direct way to express the equation of a tangent without requiring further transformations, making it especially useful for handling curves and conics.