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

Answer: The equation of the tangent line is \(y\cdot a^2 = x_0\cdot b^2 \cdot (x - x_0) + y_0\cdot a^2\).

Step-by-step solution

Differentiate the hyperbola equation

To find the slope of the tangent line, we need to differentiate the given hyperbola equation implicitly with respect to x: Given equation of hyperbola is: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) Differentiating both sides with respect to x, we get: \(\frac{2x}{a^2} - \frac{2y}{b^2}\cdot\frac{dy}{dx} = 0\) Now, let's solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2x}{a^2}\cdot\frac{b^2}{2y}\) Simplifying further: \(\frac{dy}{dx} = \frac{x\cdot b^2}{y\cdot a^2}\)

Find the slope at point (x0, y0)

Plug in the point (x0, y0) into the differentiated equation to find the specific slope at that point: \(\frac{dy}{dx} = \frac{x_0 \cdot b^2}{y_0 \cdot a^2}\) This is the slope of the tangent line at the point (x0, y0).

Write the equation of the tangent line

Use the point-slope form of the line equation: \(y - y_0 = m(x - x_0)\) Plug in the slope we found in step 2 and the point (x0, y0): \(y - y_0 = \frac{x_0 \cdot b^2}{y_0 \cdot a^2} (x - x_0)\) Now, let's rewrite the equation in a more standard form: \(y\cdot a^2 = x_0\cdot b^2 \cdot (x - x_0) + y_0\cdot a^2\) This is the equation of the line tangent to the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\).

Key concepts

Implicit Differentiation

When encountering curves defined by equations involving both x and y variables intertwined, such as a hyperbola, we employ a technique known as implicit differentiation. Unlike explicit functions, where y is expressed solely as a function of x (like y = f(x)), implicit equations involve both variables in a more complex relationship.

Implicit differentiation allows us to find the derivative of y with respect to x without solving for y explicitly. We treat y as an implicit function of x and differentiate both sides of the equation with respect to x. Remember, whenever differentiating terms with y, we add \(\frac{dy}{dx}\) since y is a function of x. For example, if differentiating \(y^2\), the result would be \(2y\frac{dy}{dx}\). This method is crucial to find the slope of the tangent line at a given point on the curve, as illustrated in the provided solution for the hyperbola.

Slope of a Tangent

The slope of a tangent line at any point on a curve is simply the value of the derivative \(\frac{dy}{dx}\) at that point. It gives us the rate at which y changes with respect to x. On a graph, this slope reflects how steep the tangent line is at the point of tangency.

In the context of curves like hyperbolas, a tangent line only touches the curve at one point, and its slope is unique to that location. To find it, we substitute the coordinates of the point of interest into the derivative we obtained through implicit differentiation. It's important to note that this slope value represents the instantaneous rate of change of the curve at that specific point, akin to the speed of a car at a precise moment in time.

Point-Slope Form

Once we have calculated the slope of the tangent line to a curve at a given point, the next step is to write down the equation of this line. The point-slope form of a linear equation is a convenient way to do this. This form is expressed as \(y - y_0 = m(x - x_0)\), where \(m\) is the slope and \(\left(x_{0}, y_{0}\right)\) is the point the line passes through.

To use this form effectively, simply substitute the slope of the tangent and the coordinates of the given point into the formula. It directly creates an equation that represents the tangent line to the curve at the point \(\left(x_{0}, y_{0}\right)\) without needing to rearrange the terms to solve for y explicitly, making it extremely handy for quickly drafting the equation of the tangent line.