HSC Extension 1 and 2 Mathematics/4-Unit/Conics

Tangent to an ellipse: Cartesian approach
The Cartesian equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$. Differentiating (using the technique of Implicit differentiation to simplify the process) to find the gradient:

Tangent to a hyperbola: Cartesian approach
The Cartesian equation of the hyperbola is $$\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$$. Differentiating (using the technique of Implicit differentiation to simplify the process) to find the gradient:
 * $$\begin{align}

0                          &= \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} \\ \frac{dy}{dx}\frac{y}{b^2} & = \frac{x}{a^2} \\ \frac{dy}{dx}              & = \frac{b^2}{a^2}\times\frac{x}{y} \end{align}$$ We can then substitute this into our point-gradient form, $$y-y_1 = m(x-x_1)$$, using the point $$P(x_1,y_1)$$:
 * at $$P$$, $$m = \frac{b^2}{a^2}\frac{x_1}{y_1}$$.
 * $$\begin{align}

y-y_1 & = \frac{b^2}{a^2}\frac{x_1}{y_1}(x-x_1) \\ yy_1-y_1^2 & = \frac{b^2}{a^2}x_1(x-x_1) \\ \frac{yy_1}{b^2}-\frac{y_1^2}{b^2} & = \frac{x_1}{a^2}(x-x_1) \\ \frac{yy_1}{b^2}-\frac{y_1^2}{b^2} & = \frac{xx_1}{a^2}-\frac{x_1^2}{a^2} \\ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2} & = \frac{xx_1}{a^2}-\frac{yy_1}{b^2} \\ \end{align}$$
 * But we know that $$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2} = 1$$ from the definition of the hyperbola, so
 * $$\frac{xx_1}{a^2}-\frac{yy_1}{b^2} = 1$$

Normal to a hyperbola: Cartesian approach
The gradient of the normal is given by $$-\frac{dx}{dy}$$, i.e., $$-\frac{a^2}{b^2}\times\frac{y}{x}$$. Finding the equation,
 * $$\begin{align}

y-y_1 & = -\frac{a^2}{b^2}\frac{y_1}{x_1}(x-x_1) \\ yb^2-y_1b^2 & = -a^2\frac{y_1}{x_1}(x-x_1) \\ \frac{b^2y}{y_1}-b^2 & = -\frac{a^2}{x_1}(x-x_1) \\ \frac{b^2y}{y_1}-b^2 & = -\frac{a^2x}{x_1}+a^2 \\ \frac{a^2x}{x_1}+\frac{b^2y}{y_1} & = a^2 + b^2 \end{align}$$