Find the equations of the tangent and normal to the hyperbola $\cfrac{{{x^2}}}{{{a^2}}} - \cfrac{{{y^2}}}{{{b^2}}} = 1$ at the point $({x_0},{y_0}).$

We have,
$\cfrac{{{x^2}}}{{{a^2}}} - \cfrac{{{y^2}}}{{{b^2}}} = 1$ …(i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{2x}}{{{a^2}}} - \cfrac{{2y\cfrac{{dy}}{{dx}}}}{{{b^2}}} = 0 \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{{b^{{2_X}}}}}{{{a^2}y}}$ …(ii)

Therefore slope of tangent, to (i) at $({x_0},{y_0})$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{({x_0},{y_0})}} = \cfrac{{{b^{}}{x_0}}}{{{a^2}{y_0}}}$

Hence, the equation of the tangent to (i) at$\;({x_0},{y_0})$ is
$y - {y_0} = \cfrac{{{b^2}{x_0}}}{{{a^2}{y_0}}}(x - {x_0})$
$\Rightarrow {a^2}{y_0}(y - {y_0}) = {b^2}{x_0}(x - {x_0})$

$\Rightarrow {b^2}x{x_0} - {a^2}y{y_0} = {b^2}x_0^2 - {a^2}y_0^2$
$\Rightarrow \cfrac{{x{x_0}}}{{{a^2}}} - \cfrac{{y{y_0}}}{{{b^2}}} = \cfrac{{x_0^2}}{{{a^2}}} - \cfrac{{y_0^2}}{{{b^2}}}$

[Dividing by ${a^2}{b^2}$]

As $({x_0},{y_0})$ lies on (i), $\cfrac{{x_0^2}}{{{a^2}}} - \cfrac{{y_0^2}}{{{b^2}}} = 1$

Hence the equation of the tangent is $\cfrac{{x{x_0}}}{{{a^2}}} - \cfrac{{y{y_0}}}{{{b^2}}} = 1.$

Slope of the normal at $({x_0},{y_0}) = \cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{tangent}}}} = - \cfrac{{{a^2}{y_0}}}{{{b^2}{x_0}}}$

Therefore the equation of the normal at $({x_0},{y_0})$
$y - {y_0} = - \cfrac{{{a^2}{y_0}}}{{{b^2}{x_0}}}(x - {x_0}) \Rightarrow \cfrac{{y - {y_0}}}{{{a^2}{y_0}}} = - \cfrac{{(x - {x_0})}}{{{b^2}{x_0}}}$

$\Rightarrow \cfrac{{y - {y_0}}}{{{a^2}{y_0}}} + \cfrac{{x - {x_0}}}{{{b^2}{x_0}}} = 0.$