Equation of Tangent to an Ellipse

The equation of a tangent to an ellipse is a linear equation that represents a line that just touches the ellipse at one point. This line is called a tangent line, and the point at which it touches the ellipse is known as the point of tangency.

Standard Form of an Ellipse

Before we derive the equation of the tangent, let's recall the standard form of an ellipse. An ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$ is given by the equation:

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

Equation of Tangent at a Point

If we have a point $(x_1, y_1)$ on the ellipse, the equation of the tangent line at this point can be written using the point-slope form:

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

This is derived from the general equation of the ellipse by applying the concept of implicit differentiation.

Slope Form of the Tangent

If the slope of the tangent line is known, say $m$, and the ellipse is in standard form, the equation of the tangent line can be written as:

$$ y = mx \pm \sqrt{a^2m^2 + b^2} $$

Parametric Form of the Tangent

For an ellipse in parametric form, where $x = a \cos \theta$ and $y = b \sin \theta$, the equation of the tangent at the point corresponding to the parameter $\theta$ is:

$$ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 $$

Differences and Important Points

Aspect	Equation of Tangent at a Point $(x_1, y_1)$	Slope Form of the Tangent	Parametric Form of the Tangent
Equation	$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$	$y = mx \pm \sqrt{a^2m^2 + b^2}$	$\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$
Known Information	Point $(x_1, y_1)$ on the ellipse	Slope $m$ of the tangent	Angle $\theta$ corresponding to the point
Implicit Differentiation	Not required	Required to find $m$	Not required
Application	When a specific point is given	When slope is given or required	When dealing with parametric equations

Examples

Example 1: Tangent at a Point

Given an ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and a point $(4, 3)$ on the ellipse, find the equation of the tangent line at this point.

Using the formula for the tangent at a point, we have:

$$ \frac{x \cdot 4}{16} + \frac{y \cdot 3}{9} = 1 \implies \frac{x}{4} + \frac{y}{3} = 1 $$

Thus, the equation of the tangent line is $x + \frac{4y}{3} = 4$.

Example 2: Slope Form of the Tangent

For the same ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the equation of the tangent line with a slope of $1$.

Using the slope form of the tangent, we get:

$$ y = x \pm \sqrt{16 + 9} \implies y = x \pm 5 $$

So there are two tangent lines with a slope of $1$, given by $y = x + 5$ and $y = x - 5$.

Example 3: Parametric Form of the Tangent

For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, find the equation of the tangent line at the point corresponding to $\theta = \frac{\pi}{3}$.

Using the parametric form of the tangent, we have:

$$ \frac{x \cos \left(\frac{\pi}{3}\right)}{4} + \frac{y \sin \left(\frac{\pi}{3}\right)}{3} = 1 \implies \frac{x \cdot \frac{1}{2}}{4} + \frac{y \cdot \frac{\sqrt{3}}{2}}{3} = 1 $$

Simplifying, we get the equation of the tangent line as $x + \sqrt{3}y = 8$.

Understanding the equation of the tangent to an ellipse is crucial for solving problems in coordinate geometry, especially in exams where questions often require finding tangent lines or proving that a given line is tangent to an ellipse.