Understanding the Equation of Normal

In geometry, particularly in the study of conic sections, the concept of a normal line is crucial. A normal to a curve at a given point is a straight line perpendicular to the tangent to the curve at that point. When we talk about the equation of the normal, we are referring to the equation that represents this perpendicular line. Let's delve deeper into this concept, particularly as it applies to the parabola.

Equation of Normal to a Parabola

A parabola is a conic section that can be defined as the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). The equation of a parabola with its vertex at the origin and axis of symmetry along the x-axis can be written as:

$$ y^2 = 4ax $$

where ( a ) is the distance from the vertex to the focus (and also to the directrix).

Deriving the Equation of Normal

To find the equation of the normal to the parabola at a point ( P(x_1, y_1) ), we need to first find the slope of the tangent at that point. The derivative of the parabola's equation gives us the slope of the tangent:

$$ \frac{dy}{dx} = \frac{d}{dx}(4ax) = 4a $$

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal at ( P ) is:

$$ m_{normal} = -\frac{1}{4a} $$

Using the point-slope form of the equation of a line, the equation of the normal at point ( P ) is:

$$ y - y_1 = m_{normal}(x - x_1) $$ $$ y - y_1 = -\frac{1}{4a}(x - x_1) $$

Simplifying, we get the equation of the normal to the parabola at ( P ):

$$ y = -\frac{1}{4a}(x - x_1) + y_1 $$

Table of Differences and Important Points

Aspect	Tangent	Normal
Definition	A line that touches the curve at a single point without crossing it.	A line perpendicular to the tangent at the point of contact.
Slope (for parabola (y^2 = 4ax))	( m_{tangent} = 4a )	( m_{normal} = -\frac{1}{4a} )
Equation (at point ( P(x_1, y_1) ))	( y = 4a(x - x_1) + y_1 )	( y = -\frac{1}{4a}(x - x_1) + y_1 )
Geometric Relation	The tangent is parallel to the axis of symmetry at the vertex.	The normal is perpendicular to the axis of symmetry at the vertex.

Examples

Example 1: Find the equation of the normal to the parabola ( y^2 = 8x ) at the point ( (2, 4) ).

Solution:

Given the parabola ( y^2 = 8x ), we have ( a = 2 ). The slope of the normal is ( m_{normal} = -\frac{1}{4a} = -\frac{1}{8} ).

Using the point-slope form:

$$ y - 4 = -\frac{1}{8}(x - 2) $$ $$ y = -\frac{1}{8}x + \frac{1}{4} + 4 $$ $$ y = -\frac{1}{8}x + \frac{17}{4} $$

Therefore, the equation of the normal is ( y = -\frac{1}{8}x + \frac{17}{4} ).

Example 2: If the normal at a point ( P ) on the parabola ( y^2 = 12x ) passes through the point ( (5, -1) ), find the coordinates of ( P ).

Solution:

For the parabola ( y^2 = 12x ), we have ( a = 3 ). The slope of the normal is ( m_{normal} = -\frac{1}{4a} = -\frac{1}{12} ).

Let ( P(x_1, y_1) ) be the point on the parabola. The equation of the normal at ( P ) is:

$$ y + 1 = -\frac{1}{12}(x - 5) $$

Since ( P ) lies on the parabola, it satisfies ( y_1^2 = 12x_1 ). Substituting ( y_1 ) in the normal equation:

$$ y_1 + 1 = -\frac{1}{12}(x_1 - 5) $$

Squaring both sides and substituting ( y_1^2 = 12x_1 ), we get a quadratic equation in ( x_1 ). Solving this equation will give us the x-coordinate of ( P ), and substituting ( x_1 ) back into ( y_1^2 = 12x_1 ) will give us the y-coordinate.

This example requires solving a quadratic equation, which is beyond the scope of this explanation, but the process outlined above is the method you would use to find the coordinates of ( P ).

Understanding the equation of the normal is essential for solving problems related to the geometry of parabolas and other conic sections. It is also a fundamental concept in calculus, where normals are used to find the shortest distance to a curve from a point outside the curve, among other applications.