Equation of Pair of Tangents

When we discuss the equation of a pair of tangents to a conic section, we are referring to the set of all lines that touch the conic at exactly one point. For an ellipse, which is the focus of this content, a pair of tangents can be drawn from a point outside the ellipse. These tangents will touch the ellipse at distinct points.

Standard Equation of an Ellipse

Before we delve into the equation of pair of tangents, let's recall the standard equation of an ellipse centered at the origin:

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

where $a$ is the semi-major axis and $b$ is the semi-minor axis.

Equation of a Tangent to an Ellipse

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of the tangent at a point $(x_1, y_1)$ on the ellipse is given by:

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

Equation of Pair of Tangents from a Point

The equation of pair of tangents drawn from a point $(x_0, y_0)$ to the ellipse can be found using the concept of homogenization. The combined equation of the pair of tangents is given by:

$$ S S_1 = T^2 $$

where $S$ is the equation of the ellipse, $S_1$ is obtained by replacing $x^2$ with $x x_0$, $y^2$ with $y y_0$, and the constant term with $-a^2 b^2$, and $T$ is the equation of the tangent in terms of point $(x_0, y_0)$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of pair of tangents from a point $(x_0, y_0)$ is:

$$ \left(\frac{x x_0}{a^2} + \frac{y y_0}{b^2}\right)^2 = \left(\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} - 1\right)\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) $$

Important Points and Differences

Point/Difference	Single Tangent	Pair of Tangents
Point of Contact	Defined (touches at a single point)	Not directly defined (touches at two distinct points)
Equation Form	Linear in $x$ and $y$	Quadratic in $x$ and $y$
Number of Tangents	One	Two
Condition for Existence	Point must lie on the ellipse	Point must lie outside the ellipse

Examples

Example 1: Single Tangent

Find the equation of the tangent to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ at the point $(3, 2)$.

Using the formula for the tangent at a point $(x_1, y_1)$:

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

we get:

$$ \frac{x \cdot 3}{9} + \frac{y \cdot 2}{4} = 1 $$

which simplifies to:

$$ \frac{x}{3} + \frac{y}{2} = 1 $$

or

$$ 2x + 3y = 6 $$

Example 2: Pair of Tangents

Find the equation of the pair of tangents drawn from the point $(4, 6)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Using the formula for the pair of tangents:

$$ \left(\frac{x x_0}{a^2} + \frac{y y_0}{b^2}\right)^2 = \left(\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} - 1\right)\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) $$

we plug in $(x_0, y_0) = (4, 6)$ and get:

$$ \left(\frac{x \cdot 4}{9} + \frac{y \cdot 6}{4}\right)^2 = \left(\frac{4^2}{9} + \frac{6^2}{4} - 1\right)\left(\frac{x^2}{9} + \frac{y^2}{4}\right) $$

which simplifies to:

$$ \left(\frac{4x}{9} + \frac{3y}{2}\right)^2 = \left(\frac{16}{9} + \frac{36}{4} - 1\right)\left(\frac{x^2}{9} + \frac{y^2}{4}\right) $$

After simplifying further, we get the quadratic equation representing the pair of tangents.

These examples illustrate how to find the equation of a single tangent and the equation of a pair of tangents to an ellipse. It's important to note that the equation of a pair of tangents is more complex and requires a point outside the ellipse, while a single tangent requires a point on the ellipse.