Q.Find the equation of the tangent to the hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ at the point in the first quadrant whose ordinate is 3.

Differentiating the standard hyperbola equation gives slope $dy/dx=\dfrac{b^2x}{a^2y}$, leading to the point-form tangent at $(x_1,y_1)$: $\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1$, and — using the parametric point — the eccentric-angle form $\dfrac{x}{a}\sec\theta_1-\dfrac{y}{b}\tan\theta_1=1$.

The condition of tangency for a line $y=mx+c$ works out to $c^2=a^2m^2-b^2$ — the one place the hyperbola's formulas differ in SIGN from the ellipse's analogous $c^2=a^2m^2+b^2$. A consequence worth noting: because the right side must be non-negative for $c$ to be real, tangent lines to a hyperbola can only have slopes $m$ satisfying $a^2m^2\ge b^2$ — unlike the ellipse, which admits a tangent of every possible slope. The point of contact is $\left(-\dfrac{a^2m}{c},-\dfrac{b^2}{c}\right)$ — both coordinates carry a minus sign, again the mirror of the ellipse's point of contact. …