Tangent at a Point on the Ellipse

Tangent to x²/a² + y²/b² = 1 at point (x₁, y₁) lying on it. Obtained by the T = 0 rule: replace x² → xx₁, y² → yy₁.

Let $P(x_1, y_1)$ lie on $x^2/a^2 + y^2/b^2 = 1$, so $x_1^2/a^2 + y_1^2/b^2 = 1$.

Differentiating implicitly:

$$\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$$

Slope at $P$: $m = -b^2x_1/(a^2y_1)$.

Tangent through $P(x_1, y_1)$:

$$y - y_1 = -\frac{b^2x_1}{a^2y_1}(x - x_1)$$ $$a^2y_1(y-y_1) = -b^2x_1(x-x_1)$$ $$a^2y_1y - a^2y_1^2 = -b^2x_1x + b^2x_1^2$$ $$b^2x_1x + a^2y_1y = b^2x_1^2 + a^2y_1^2 = a^2b^2$$

Dividing by $a^2b^2$:

$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$

The T = 0 rule: Replace $x^2 \to xx_1$, $y^2 \to yy_1$ in the ellipse equation. This pattern is universal across all standard conics.