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:

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

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 $P$, $m={\frac {b^{2}}{a^{2}}}{\frac {x_{1}}{y_{1}}}$.

${\begin{aligned}y-y_{1}&={\frac {b^{2}}{a^{2}}}{\frac {x_{1}}{y_{1}}}(x-x_{1})\\yy_{1}-y_{1}^{2}&={\frac {b^{2}}{a^{2}}}x_{1}(x-x_{1})\\{\frac {yy_{1}}{b^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {x_{1}}{a^{2}}}(x-x_{1})\\{\frac {yy_{1}}{b^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {xx_{1}}{a^{2}}}-{\frac {x_{1}^{2}}{a^{2}}}\\{\frac {x_{1}^{2}}{a^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {xx_{1}}{a^{2}}}-{\frac {yy_{1}}{b^{2}}}\\\end{aligned}}$

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,

${\begin{aligned}y-y_{1}&=-{\frac {a^{2}}{b^{2}}}{\frac {y_{1}}{x_{1}}}(x-x_{1})\\yb^{2}-y_{1}b^{2}&=-a^{2}{\frac {y_{1}}{x_{1}}}(x-x_{1})\\{\frac {b^{2}y}{y_{1}}}-b^{2}&=-{\frac {a^{2}}{x_{1}}}(x-x_{1})\\{\frac {b^{2}y}{y_{1}}}-b^{2}&=-{\frac {a^{2}x}{x_{1}}}+a^{2}\\{\frac {a^{2}x}{x_{1}}}+{\frac {b^{2}y}{y_{1}}}&=a^{2}+b^{2}\end{aligned}}$