Equation of normal
Equation of Normal
In geometry, the normal to a curve at a particular point is the line perpendicular to the tangent to the curve at that point. When dealing with conic sections such as ellipses, the equation of the normal line can be derived using the properties of the ellipse and the slope of the tangent line at the point of interest.
Equation of an Ellipse
An ellipse is defined as the set of all points in a plane the sum of whose distances from two fixed points (foci) is constant. The standard form of the equation of an ellipse centered at the origin with its major axis along the x-axis is given by:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
where (a) is the semi-major axis and (b) is the semi-minor axis.
Slope of the Tangent Line
The slope of the tangent line to the ellipse at any point ((x_1, y_1)) can be found by differentiating the equation of the ellipse with respect to (x):
$$ \frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{d}{dx}(1) \ \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 $$
Solving for (y'), the derivative of (y) with respect to (x), gives us the slope of the tangent line (m_t):
$$ y' = m_t = -\frac{b^2x_1}{a^2y_1} $$
Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, if (m_t) is the slope of the tangent, the slope of the normal line (m_n) is:
$$ m_n = -\frac{1}{m_t} = \frac{a^2y_1}{b^2x_1} $$
The equation of the normal line at the point ((x_1, y_1)) on the ellipse can be written in point-slope form:
$$ y - y_1 = m_n(x - x_1) $$
Substituting the value of (m_n), we get:
$$ y - y_1 = \frac{a^2y_1}{b^2x_1}(x - x_1) $$
Differences and Important Points
| Aspect | Tangent Line | Normal Line |
|---|---|---|
| Definition | Line that touches the curve at one point | Line perpendicular to the tangent at that point |
| Slope ((m)) | Given by (-\frac{b^2x_1}{a^2y_1}) | Given by (\frac{a^2y_1}{b^2x_1}) |
| Equation Form | (y - y_1 = m_t(x - x_1)) | (y - y_1 = m_n(x - x_1)) |
| Relationship | (m_t \cdot m_n = -1) | (m_t \cdot m_n = -1) |
| Application | Used to find the tangent at a point | Used to find the perpendicular at a point |
Examples
Example 1: Find the equation of the normal to the ellipse (\frac{x^2}{16} + \frac{y^2}{9} = 1) at the point ((2, \frac{3}{2})).
- First, find the slope of the tangent line using the derivative:
$$ m_t = -\frac{b^2x_1}{a^2y_1} = -\frac{9 \cdot 2}{16 \cdot \frac{3}{2}} = -\frac{3}{4} $$
- Then, find the slope of the normal line:
$$ m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} $$
- Finally, write the equation of the normal line using the point-slope form:
$$ y - \frac{3}{2} = \frac{4}{3}(x - 2) $$
Example 2: Determine the equation of the normal to the ellipse (\frac{x^2}{25} + \frac{y^2}{16} = 1) at the point ((3, 4)).
- Calculate the slope of the tangent line:
$$ m_t = -\frac{b^2x_1}{a^2y_1} = -\frac{16 \cdot 3}{25 \cdot 4} = -\frac{12}{25} $$
- Find the slope of the normal line:
$$ m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{12}{25}} = \frac{25}{12} $$
- Write the equation of the normal line:
$$ y - 4 = \frac{25}{12}(x - 3) $$
These examples illustrate how to find the equation of the normal to an ellipse at a given point. It's important to remember that the normal is perpendicular to the tangent and that its slope is the negative reciprocal of the slope of the tangent.