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:

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

We can then substitute this into our point-gradient form, y − y₁ = m(x − x₁), using the point P(x₁,y₁):

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,

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