Equation of Tangent

In mathematics, particularly in calculus and analytic geometry, the equation of a tangent line to a curve at a given point is a straight line that just touches the curve at that point. This line is perpendicular to the normal line to the curve at the same point. The concept of a tangent is essential in many areas of mathematics, including differential calculus and geometric constructions.

Equation of a Tangent Line

The general equation of a tangent line to a curve defined by a function $f(x)$ at a point $x = a$ is given by:

$$ y - f(a) = f'(a)(x - a) $$

where $f'(a)$ is the derivative of the function $f(x)$ at the point $x = a$. This derivative represents the slope of the tangent line at that point.

Tangent to a Hyperbola

A hyperbola is a type of conic section that can be defined as the set of all points $(x, y)$ in the plane such that the difference of the distances from two fixed points (foci) is constant. The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is:

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$

The equation of the tangent line to this hyperbola at a point $(x_1, y_1)$ on the hyperbola is given by:

$$ \frac{x_1x}{a^2} - \frac{y_1y}{b^2} = 1 $$

Point-Slope Form

If you have a point $(x_1, y_1)$ on the hyperbola and the slope $m$ of the tangent line at that point, the equation of the tangent line can also be written in point-slope form:

$$ y - y_1 = m(x - x_1) $$

Slope of the Tangent to a Hyperbola

The slope of the tangent line to the hyperbola at the point $(x_1, y_1)$ can be found by differentiating the standard equation of the hyperbola with respect to $x$:

$$ \frac{d}{dx}\left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = \frac{d}{dx}(1) $$

This gives us:

$$ \frac{2x}{a^2} - \frac{2yy'}{b^2} = 0 $$

Solving for $y'$ (the slope of the tangent line) gives:

$$ y' = \frac{b^2x}{a^2y} $$

Differences and Important Points

Here is a table summarizing the differences and important points regarding the equation of a tangent line to a hyperbola:

Aspect	Description
Standard Equation	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ for a hyperbola centered at the origin
Equation of Tangent	$\frac{x_1x}{a^2} - \frac{y_1y}{b^2} = 1$ for a point $(x_1, y_1)$ on the hyperbola
Slope (m)	$m = \frac{b^2x_1}{a^2y_1}$ for the slope of the tangent line at $(x_1, y_1)$
Point-Slope Form	$y - y_1 = m(x - x_1)$ for the tangent line at $(x_1, y_1)$ with slope $m$

Examples

Example 1: Find the Equation of the Tangent Line

Find the equation of the tangent line to the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ at the point $(3, 4)$.

Solution:

Using the formula for the equation of the tangent line to a hyperbola, we have:

$$ \frac{x_1x}{a^2} - \frac{y_1y}{b^2} = 1 $$

Plugging in the values $x_1 = 3$, $y_1 = 4$, $a^2 = 9$, and $b^2 = 16$, we get:

$$ \frac{3x}{9} - \frac{4y}{16} = 1 $$

Simplifying, we obtain the equation of the tangent line:

$$ x - \frac{y}{4} = 3 $$

Example 2: Using the Slope to Find the Tangent Line

Given the hyperbola $\frac{x^2}{25} - \frac{y^2}{9} = 1$, find the equation of the tangent line at the point $(5, 3)$.

Solution:

First, we find the slope of the tangent line using the formula:

$$ m = \frac{b^2x_1}{a^2y_1} $$

Plugging in the values $x_1 = 5$, $y_1 = 3$, $a^2 = 25$, and $b^2 = 9$, we get:

$$ m = \frac{9 \cdot 5}{25 \cdot 3} = \frac{3}{5} $$

Now, using the point-slope form of the equation of the tangent line, we have:

$$ y - 3 = \frac{3}{5}(x - 5) $$

Expanding and rearranging, we get the equation of the tangent line:

$$ 5y - 15 = 3x - 15 $$

$$ 3x - 5y = 0 $$

These examples illustrate how to find the equation of the tangent line to a hyperbola at a given point using different methods. Understanding these concepts is crucial for solving problems related to tangents and hyperbolas in exams.