Tangent to an ellipse: Cartesian approach

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

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

We can then substitute this into our point-gradient form, ${\displaystyle y-y_{1}=m(x-x_{1})}$, using the point ${\displaystyle P(x_{1},y_{1})}$:

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

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

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

${\displaystyle {\frac {xx_{1}}{a^{2}}}-{\frac {yy_{1}}{b^{2}}}=1}$

Normal to a hyperbola: Cartesian approach

The gradient of the normal is given by ${\displaystyle -{\frac {dx}{dy}}}$, i.e., ${\displaystyle -{\frac {a^{2}}{b^{2}}}\times {\frac {y}{x}}}$. Finding the equation,

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