Equation of Tangent to Ellipse in Reduced Form

Theorem
Let $E$ be an ellipse embedded in a Cartesian plane in reduced form with the equation:
 * $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

Let $P = \tuple {x_1, y_1}$ be a point on $E$.

The tangent to $E$ at $P$ is given by the equation:
 * $\dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2} = 1$

Proof
From the slope-intercept form of a line, the equation of a line passing through $P$ is:
 * $y - y_1 = \mu \paren {x - x_1}$

If this line passes through another point $\tuple {x_2, y_2}$ on $E$, the slope of the line is given by:
 * $\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$

Because $P$ and $Q$ both lie on $E$, we have:

As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.

The limit of the slope is therefore:


 * $-\dfrac {2 b^2 x_1} {2 a^2 y_1} = -\dfrac {b^2 x_1} {a^2 y_1}$

The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore: