Welcome to this module on Parametric Equations of an Ellipse, a crucial topic for competitive exams like JEE. Understanding parametric forms allows us to represent points on an ellipse using a single variable, simplifying many calculations and problem-solving approaches.

Parametric equations express the coordinates of points on a curve as functions of an independent variable, called the parameter. For an ellipse, this parameter is typically an angle.

Consider the standard equation of an ellipse centered at the origin with semi-major axis 'a' along the x-axis and semi-minor axis 'b' along the y-axis:

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

We can represent any point $(x, y)$ on this ellipse using the parameter $\theta$ (theta), which represents the angle made by the line joining the origin to the point $(x, y)$ with the positive x-axis. The parametric equations are:

$x = a \cos \theta$ $y = b \sin \theta$

As $\theta$ varies from $0$ to $2\pi$, the point $(x, y)$ traces out the entire ellipse.

If the ellipse is centered at $(h, k)$, its equation becomes:

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

The parametric equations are then modified to account for the translation:

$x = h + a \cos \theta$ $y = k + b \sin \theta$

Here, $\theta$ still represents the eccentric anomaly.

Parametric equations are incredibly useful for:

• Finding points on the ellipse: Simply substitute a value for $\theta$.
• Calculating tangents and normals: The derivatives of the parametric equations simplify these calculations.
• Determining area and perimeter: Integration with respect to the parameter is often easier.
• Solving locus problems: Representing points parametrically can simplify complex locus derivations.

Find the equation of the tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at the point whose eccentric anomaly is $\frac{\pi}{3}$.

First, find the coordinates of the point: $x = 4 \cos(\frac{\pi}{3}) = 4 \times \frac{1}{2} = 2$ $y = 3 \sin(\frac{\pi}{3}) = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$

The point is $(2, \frac{3\sqrt{3}}{2})$.

The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$.

Substituting the values: $\frac{x}{4} \cos(\frac{\pi}{3}) + \frac{y}{3} \sin(\frac{\pi}{3}) = 1$ $\frac{x}{4} \times \frac{1}{2} + \frac{y}{3} \times \frac{\sqrt{3}}{2} = 1$ $\frac{x}{8} + \frac{\sqrt{3}y}{6} = 1$

Multiplying by 24 to clear denominators: $3x + 4\sqrt{3}y = 24$.