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

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Ellipses[edit]

Tangent to an ellipse: Cartesian approach[edit]

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[edit]

Tangent to a hyperbola: Cartesian approach[edit]

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[edit]

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}