Equation of Chord of Contact on Ellipse in Reduced Form

Theorem
Let $\EE$ 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 $P = \tuple {x_0, y_0}$ be a point which is outside the boundary of $\EE$.

Let $UV$ be the chord of contact on $\EE$ with respect to $P$.

Then $UV$ can be defined by the equation:


 * $\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$

Proof
Let $\TT_1$ and $\TT_2$ be a tangents to $\EE$ passing through $P$.

Let:
 * $\TT_1$ touch $\EE$ at $U = \tuple {x_1, y_1}$
 * $\TT_2$ touch $\EE$ at $V = \tuple {x_2, y_2}$

Then the chord of contact on $\EE$ with respect to $P$ is defined as $UV$.


 * Equation-of-polar-of-ellipse.png

From Equation of Tangent to Ellipse in Reduced Form, $\TT_1$ is expressed by the equation:


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

but as $\TT_1$ also passes through $\tuple {x_0, y_0}$ we also have:


 * $\dfrac {x_0 x_1} {a^2} + \dfrac {y_0 y_1} {b^2} = 1$

This also expresses the condition that $U$ should lie on $\TT_1$:
 * $\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$

Similarly, From Equation of Tangent to Circle Centered at Origin, $\TT_2$ is expressed by the equation:


 * $\dfrac {x x_2} {a^2} + \dfrac {y y_2} {b^2} = 1$

but as $\TT_2$ also passes through $\tuple {x_0, y_0}$ we also have:


 * $\dfrac {x_0 x_2} {a^2} + \dfrac {y_0 y_2} {b^2} = 1$

This also expresses the condition that $V$ should lie on $\TT_2$:
 * $\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$

So both $U$ and $V$ lie on the straight line whose equation is:
 * $\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$

and the result follows.