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Show that the equation of the tangent line to. $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$ at the point \(\left(x_{0}, y_{0}\right)\) is \(\left(x_{0} / a^{2}\right) x-\left(y_{0} / b^{2}\right) y=1\).

Short Answer

Expert verified
The equation of the tangent line to the hyperbola at a given point is indeed \(\left(\frac{x_0}{a^2}\right) x - \left(\frac{y_0}{b^2}\right) y = 1\) as verified by calculus and algebraic manipulations.

Step by step solution

01

Derive the equation

Start by taking the derivative of the equation of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\). With \(y\) as a function of \(x\), implicit differentiation will be used. The derivative \(dy/dx\) is then given by \(\frac{dy}{dx} = \frac{xb^2}{ya^2}\). Note that this will provide the slope of the tangent line at any given point on the hyperbola.
02

Compute the slope of tangent line

Substitute the point \((x_{0}, y_{0})\) into the slope equation to find the slope at the specified point. This will be \(\frac{x_0b^2}{y_0a^2}\).
03

Formulate the equation of the tangent line

Use the point-slope form of a line (which is \(y-y_1=m(x-x_1)\) where \(m\) is the slope and \((x_1, y_1)\) is a point on the line) to formulate the equation of the tangent line. Substitute the slope we found earlier and the given point into the point-slope form, and simplify to obtain: \(\frac{x_0}{a^2}x - \frac{y_0}{b^2}y = 1\).

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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 process utilized in calculus to find the derivative of equations where the dependent and independent variables are intermixed. In many cases, equations will not be explicitly solved for a particular variable, so this technique allows us to differentiate implicitly.
For our hyperbola equation, \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), both \(x\) and \(y\) are within the equation. We assume \(y\) is a function of \(x\), but we don't solve for \(y\) explicitly. Instead, we differentiate each term with respect to \(x\), while applying the chain rule to \(dy/dx\) where needed.
This means for each \(y\)-related term, we differentiate as if \(y\) is a function of \(x\), multiplying by \(dy/dx\) as part of the process. For example, differentiating \(-\frac{y^2}{b^2}\) would produce \(-\frac{2y}{b^2} \times \frac{dy}{dx}\). Join these results to form a derivative equation that provides valuable information like the slope of the tangent.
Equation of a Hyperbola
A hyperbola is a type of conic section that appears as two open curves, mirroring each other. The standard form for a hyperbola aligned along the x and y-axes is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
The equation represents all the points \((x, y)\) that form the shape of the hyperbola. As a tool, this formula is crucial for problems involving hyperbolas, such as finding the tangent line, or assessing how changes in \(a\) and \(b\) affect the graph's orientation and spread. To verify that a tangent line meets the hyperbola at only one point, this equation is instrumental.
In our original exercise, moving forward with this hyperbola's equation sets the stage for further operations like differentiation and, ultimately, identifying the tangent line.
Point-Slope Form
The point-slope form of a line is a reliable method for writing the equation of a line when you know a point on the line and the slope. It is given by the formula: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is a specific point on the line.
In the context of finding a tangent line to a hyperbola, utilizing the point-slope form simplifies the process. Knowing the slope of the tangent (obtained through implicit differentiation) and the point of tangency \((x_0, y_0)\), we insert these values into the point-slope formula.
After filling in the known values, the equation can be rearranged into another form, such as the standard linear equation \(\frac{x_0}{a^2}x - \frac{y_0}{b^2}y = 1\).
Mastery of the point-slope form is a fantastic way to streamline tasks requiring tangent lines or any line equations constructed quickly around known data points.

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Most popular questions from this chapter

Show that the polar equation for \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is \(r^{2}=\frac{-b^{2}}{1-e^{2} \cos ^{2} \theta} \cdot\)

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