Problems Based on Normal

In the context of parabolas and other conic sections, the "normal" refers to a line that is perpendicular to the tangent at the point of tangency on the curve. Understanding the properties of normals is crucial for solving various problems in coordinate geometry, especially in the study of parabolas.

Properties of Normal to a Parabola

Before diving into problems, let's establish some important properties of normals to a parabola:

• A normal to a parabola is a line that is perpendicular to the tangent at the point of contact.
• For a standard parabola $y^2 = 4ax$, the slope of the normal at any point $(x_1, y_1)$ on the parabola is $-\frac{1}{2a}x_1$.
• The equation of the normal in point-slope form is given by $y - y_1 = -\frac{1}{2a}x_1(x - x_1)$.

Equation of Normal

For a parabola $y^2 = 4ax$, the equation of the normal at a point $(at^2, 2at)$ can be derived using the slope mentioned above. The equation is:

[ y = -tx + 2at + at^3 ]

This equation can be used to find various properties of the normal, such as its point of intersection with the axis of the parabola, or with any other line or curve.

Table of Differences and Important Points

Property	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 $y^2 = 4ax$)	$m = \frac{1}{2a}y_1$	$m = -\frac{1}{2a}x_1$
Equation (at point $(at^2, 2at)$)	$y = tx + at^2$	$y = -tx + 2at + at^3$
Intersection with Axis	Passes through the focus $(a, 0)$	Passes through the point $(0, 2at + at^3)$ on the y-axis

Examples

Example 1: Finding the Equation of the Normal

Problem: 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 at the point $(2, 4)$ is $-\frac{1}{2a}x_1 = -\frac{1}{4}(2) = -\frac{1}{2}$.

Using the point-slope form of the line equation:

[ y - y_1 = m(x - x_1) ]

[ y - 4 = -\frac{1}{2}(x - 2) ]

[ y = -\frac{1}{2}x + 5 ]

So, the equation of the normal is $y = -\frac{1}{2}x + 5$.

Example 2: Intersection with the Axis

Problem: Find where the normal at the point $(1, 2\sqrt{2})$ on the parabola $y^2 = 8x$ intersects the y-axis.

Solution:

For the parabola $y^2 = 8x$, we have $a = 2$. The point $(1, 2\sqrt{2})$ corresponds to the parameter $t = \sqrt{2}$.

Using the normal equation $y = -tx + 2at + at^3$, we get:

[ y = -\sqrt{2}x + 4\sqrt{2} + 2\sqrt{2} ]

To find the intersection with the y-axis, set $x = 0$:

[ y = 4\sqrt{2} + 2\sqrt{2} = 6\sqrt{2} ]

Therefore, the normal intersects the y-axis at $(0, 6\sqrt{2})$.

Example 3: Normal as a Locus

Problem: Show that the locus of the foot of the perpendicular drawn from the focus of the parabola $y^2 = 4ax$ to any normal is a straight line.

Solution:

Let the normal be at the point $(at^2, 2at)$ on the parabola. The focus of the parabola is $(a, 0)$. The equation of the normal is $y = -tx + 2at + at^3$.

The foot of the perpendicular from the focus to the normal will satisfy both the equation of the normal and the condition that the line joining the focus to the foot is perpendicular to the normal. This means that the slope of the line joining the focus to the foot is the negative reciprocal of the slope of the normal.

Let the foot of the perpendicular be $(x_0, y_0)$. The slope of the line joining the focus to the foot is:

[ m = \frac{y_0}{x_0 - a} ]

Since it is perpendicular to the normal, we have:

[ m \cdot (-t) = -1 ]

[ \frac{y_0}{x_0 - a} \cdot (-t) = -1 ]

Substituting $y_0$ from the normal equation:

[ \frac{-tx_0 + 2at + at^3}{x_0 - a} \cdot (-t) = -1 ]

Simplifying, we find that $x_0$ is eliminated, and we are left with a relation in $y_0$ and $t$ that does not involve $x_0$. This means that for any value of $t$, the $y_0$ coordinate of the foot of the perpendicular is the same, indicating that the locus is a straight line parallel to the x-axis.

In conclusion, problems based on normals to a parabola involve understanding the properties of normals, their equations, and how they interact with other geometric entities. By using the properties and equations discussed above, a wide range of problems can be solved.