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

Ellipses

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:

Hyperbolae

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:

$\begin{array}{cc}0& =\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}\\ \frac{dy}{dx}\frac{y}{b^2}& =\frac{x}{a^2}\\ \frac{dy}{dx}& =\frac{b^2}{a^2}\times \frac{x}{y}\end{array}$

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 $P$, $m=\frac{b^2}{a^2}\frac{x_1}{y_1}$.

$\begin{array}{cc}y-y_1& =\frac{b^2}{a^2}\frac{x_1}{y_1}(x-x_1)\\ y{y}_1-y_1^2& =\frac{b^2}{a^2}{x}_1(x-x_1)\\ \frac{y{y}_1}{b^2}-\frac{y_1^2}{b^2}& =\frac{x_1}{a^2}(x-x_1)\\ \frac{y{y}_1}{b^2}-\frac{y_1^2}{b^2}& =\frac{x{x}_1}{a^2}-\frac{x_1^2}{a^2}\\ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}& =\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}\\ \end{array}$

But we know that $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$ from the definition of the hyperbola, so

$\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$

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,

$\begin{array}{cc}y-y_1& =-\frac{a^2}{b^2}\frac{y_1}{x_1}(x-x_1)\\ y{b}^2-y_1{b}^2& =-a^2\frac{y_1}{x_1}(x-x_1)\\ \frac{b^{2}y}{y_1}-b^2& =-\frac{a^2}{x_1}(x-x_1)\\ \frac{b^{2}y}{y_1}-b^2& =-\frac{a^{2}x}{x_1}+a^2\\ \frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}& =a^2+b^2\end{array}$