Tangent at a Point on the Parabola

Tangent to y² = 4ax at the point (x₁, y₁) lying on it. Obtained by the T = 0 rule: replace y² → yy₁, x → (x+x₁)/2.

Let $P(x_1, y_1)$ lie on $y^2 = 4ax$, so $y_1^2 = 4ax_1$.

Differentiating $y^2 = 4ax$ implicitly:

$$2y\frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}$$

Slope of tangent at $P$: $m = \dfrac{2a}{y_1}$.

Equation of tangent through $P(x_1, y_1)$:

$$y - y_1 = \frac{2a}{y_1}(x - x_1)$$ $$y_1(y - y_1) = 2a(x - x_1)$$ $$yy_1 - y_1^2 = 2ax - 2ax_1$$

Since $y_1^2 = 4ax_1$:

$$yy_1 - 4ax_1 = 2ax - 2ax_1$$ $$yy_1 = 2ax + 2ax_1 = 2a(x + x_1)$$

The T = 0 substitution rule: In $y^2 = 4ax$, replace $y^2 \to yy_1$ and $x \to \frac{x+x_1}{2}$ to get the tangent at $(x_1, y_1)$. This rule extends to all standard conics.