Tangent and Normal at a Point — Cartesian Form

The tangent and normal at a point on a hyperbola are built exactly the way they are for an ellipse or parabola — differentiate implicitly, or simply quote the standard result, which is easiest to remember using the $S_1$ shorthand introduced earlier.

Tangent at $P(x_1,y_1)$. The equation of the tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ at the point $P(x_1,y_1)$ lying on the curve is obtained by "replacing $x^2$ with $xx_1$ and $y^2$ with $yy_1$" in the equation — the same half-and-half substitution rule used across all conics. In the $S_1$ notation this is simply

Normal at $P(x_1,y_1)$. The normal is the line through $P$ perpendicular to the tangent there. Its equation works out to

The restriction $y_1\neq 0$ just excludes the vertices $(\pm a, 0)$, where the tangent is vertical and the normal is simply the $x$-axis itself — a degenerate case that doesn't fit the general formula's denominators.

Worked example. Take the point $P=(6, 2\sqrt3)$ on $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ (the same point as $\theta=60^\circ$ from the previous section).

Tangent: substituting $x_1=6,y_1=2\sqrt3,a^2=9,b^2=4$ into $S_1=0$:

Multiplying through by $6$ to clear denominators: $4x - 3\sqrt3\,y = 6$.

Normal: substituting into $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$:

$\dfrac{9x}{6} + \dfrac{4y}{2\sqrt3} = 13 \ \Longrightarrow\ \dfrac{3x}{2} + \dfrac{2y}{\sqrt3} = 13.$ …