Equation of Normal

In geometry, the normal to a curve at a particular point is a line perpendicular to the tangent to the curve at that point. When dealing with conic sections such as hyperbolas, the equation of the normal line can be derived using calculus or geometric properties.

Hyperbola

A hyperbola is a type of conic section that is formed by the intersection of a plane and a double cone, where the plane cuts through both halves of the cone. The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is given by:

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

where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively.

Equation of Tangent

Before we discuss the equation of the normal, it is important to understand the equation of the tangent to a hyperbola. At a point $(x_1, y_1)$ on the hyperbola, the equation of the tangent line is:

$$ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 $$

Equation of Normal

The equation of the normal to the hyperbola at the point $(x_1, y_1)$ is derived by finding the line perpendicular to the tangent at that point. The slope of the tangent line is the negative reciprocal of the slope of the normal line.

Given the slope of the tangent line to the hyperbola at $(x_1, y_1)$ is:

$$ m_{\text{tangent}} = \frac{dy}{dx} = \frac{b^2 x_1}{a^2 y_1} $$

The slope of the normal line, $m_{\text{normal}}$, is:

$$ m_{\text{normal}} = -\frac{a^2 y_1}{b^2 x_1} $$

Using the point-slope form of the equation of a line, the equation of the normal at $(x_1, y_1)$ is:

$$ y - y_1 = m_{\text{normal}}(x - x_1) $$

Substituting the value of $m_{\text{normal}}$:

$$ y - y_1 = -\frac{a^2 y_1}{b^2 x_1}(x - x_1) $$

Simplifying, we get the equation of the normal to the hyperbola:

$$ b^2 x_1 (y - y_1) + a^2 y_1 (x - x_1) = 0 $$

Or, in the standard form:

$$ b^2 x_1 y + a^2 y_1 x = (a^2 + b^2) x_1 y_1 $$

Differences and Important Points

Aspect	Tangent Line	Normal Line
Definition	Line that touches the curve at one point	Line perpendicular to the tangent at a point
Slope (Hyperbola)	$m_{\text{tangent}} = \frac{b^2 x_1}{a^2 y_1}$	$m_{\text{normal}} = -\frac{a^2 y_1}{b^2 x_1}$
Equation (Hyperbola)	$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$	$b^2 x_1 y + a^2 y_1 x = (a^2 + b^2) x_1 y_1$
Geometric Relation	Touches the curve	Perpendicular to the tangent

Examples

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

Given $a^2 = 16$ and $b^2 = 9$, and the point $(x_1, y_1) = (4, 3)$, we can use the equation of the normal:

$$ b^2 x_1 y + a^2 y_1 x = (a^2 + b^2) x_1 y_1 $$

Substituting the values:

$$ 9 \cdot 4 \cdot y + 16 \cdot 3 \cdot x = (16 + 9) \cdot 4 \cdot 3 $$

Simplifying:

$$ 36y + 48x = 100 \cdot 12 $$

$$ 3y + 4x = 100 $$

This is the equation of the normal to the given hyperbola at the point $(4, 3)$.

Example 2: If the normal at a point $(x_1, y_1)$ on the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ passes through the point $(12, -5)$, find $(x_1, y_1)$.

Using the equation of the normal:

$$ b^2 x_1 y + a^2 y_1 x = (a^2 + b^2) x_1 y_1 $$

Substitute $a^2 = 25$, $b^2 = 16$, and the point $(12, -5)$:

$$ 16 x_1 (-5) + 25 y_1 (12) = (25 + 16) x_1 y_1 $$

Solving for $x_1$ and $y_1$:

$$ -80 x_1 + 300 y_1 = 41 x_1 y_1 $$

This equation can be solved along with the hyperbola equation to find the point $(x_1, y_1)$ where the normal passes through $(12, -5)$. This typically requires solving a system of nonlinear equations, which may have more than one solution depending on the context.

Understanding the equation of the normal is crucial for various applications in geometry, physics, and engineering, where the concept of perpendicularity to a curve is used.