Chord with a Given Midpoint

Chord of the ellipse whose midpoint is (x₁, y₁). Explicitly: xx₁/a² + yy₁/b² − 1 = x₁²/a² + y₁²/b² − 1, i.e. T = S₁.

Let $(x_1, y_1)$ be the midpoint of a chord of $x^2/a^2 + y^2/b^2 = 1$ with endpoints $(x_2, y_2)$ and $(x_3, y_3)$.

Since both endpoints are on the ellipse:

$$\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} = 1 \quad \text{and} \quad \frac{x_3^2}{a^2} + \frac{y_3^2}{b^2} = 1$$

Subtracting:

$$\frac{x_2^2 - x_3^2}{a^2} + \frac{y_2^2 - y_3^2}{b^2} = 0$$ $$\frac{(x_2+x_3)(x_2-x_3)}{a^2} = -\frac{(y_2+y_3)(y_2-y_3)}{b^2}$$

Using midpoint: $x_2+x_3 = 2x_1$, $y_2+y_3 = 2y_1$:

$$\frac{2x_1}{a^2} \cdot (x_2-x_3) = -\frac{2y_1}{b^2} \cdot (y_2-y_3)$$

Slope of chord: $\dfrac{y_2-y_3}{x_2-x_3} = -\dfrac{b^2x_1}{a^2y_1}$

Chord through $(x_1,y_1)$ with this slope:

$$y - y_1 = -\frac{b^2x_1}{a^2y_1}(x-x_1)$$

Rearranging (using $x_1^2/a^2 + y_1^2/b^2 = S_1 + 1$):

$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$$ $$T = S_1$$