Basing on Director Circle of Ellipse

The director circle of an ellipse is a special circle associated with the ellipse. It is defined as the locus of points from which tangents drawn to the ellipse are perpendicular to each other. In this in-depth content, we will explore the concept of the director circle of an ellipse, its properties, and how it differs from the ellipse itself.

Definition of Ellipse

Before we delve into the director circle, let's define an ellipse. An ellipse is the set of all points in a plane for which the sum of the distances to two fixed points (foci) is constant. The standard equation of an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$ is given by:

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

Director Circle of an Ellipse

The director circle of an ellipse is defined by the equation:

$$ x^2 + y^2 = a^2 + b^2 $$

This circle has a radius equal to the square root of the sum of the squares of the semi-major and semi-minor axes of the ellipse.

Properties of the Director Circle

• The director circle is concentric with the ellipse.
• The radius of the director circle is greater than the semi-major axis of the ellipse.
• Every tangent to the director circle is a chord of the ellipse that subtends a right angle at the center.

Differences between Ellipse and Director Circle

Here is a table summarizing the differences between an ellipse and its director circle:

Feature	Ellipse	Director Circle
Equation	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$x^2 + y^2 = a^2 + b^2$
Shape	Oval	Circle
Center	Same as ellipse (origin if centered at origin)	Same as ellipse
Radius	Not applicable (has semi-major and semi-minor axes)	$\sqrt{a^2 + b^2}$
Tangents	Not necessarily perpendicular	Perpendicular if they intersect on the director circle

Examples

Example 1: Finding the Director Circle

Given an ellipse with the equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the equation of its director circle.

Solution:

The semi-major axis $a$ is $\sqrt{16} = 4$, and the semi-minor axis $b$ is $\sqrt{9} = 3$. The equation of the director circle is:

$$ x^2 + y^2 = a^2 + b^2 $$

Substituting the values of $a$ and $b$, we get:

$$ x^2 + y^2 = 4^2 + 3^2 = 16 + 9 = 25 $$

Therefore, the equation of the director circle is:

$$ x^2 + y^2 = 25 $$

Example 2: Tangents from a Point on the Director Circle

Find the tangents from the point $(5, 0)$ on the director circle $x^2 + y^2 = 25$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Solution:

Since the point $(5, 0)$ lies on the director circle, the tangents from this point to the ellipse will be perpendicular to each other. To find the equations of the tangents, we use the point form of the tangent to an ellipse:

$$ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 $$

Substituting $x_1 = 5$ and $y_1 = 0$, we get:

$$ \frac{5x}{16} + \frac{0 \cdot y}{9} = 1 $$

Simplifying, we find one tangent:

$$ x = \frac{16}{5} $$

For the other tangent, we need to consider the symmetry of the ellipse. Since the point $(5, 0)$ is on the x-axis, the other tangent will be parallel to the y-axis and pass through $(5, 0)$. Therefore, the second tangent is:

$$ x = 5 $$

In conclusion, the director circle of an ellipse provides a geometric locus from which tangents to the ellipse are perpendicular. Understanding the properties and equations of the director circle is essential for solving problems related to tangents and normals of ellipses.