Tangent and Normal at 'θ', and the Condition for Tangency

The same tangent and normal lines can also be written directly in terms of the parameter $\theta$, without first computing the point's Cartesian coordinates — genuinely convenient when a problem is already phrased in terms of $\theta$.

Tangent at 'θ'.

Normal at 'θ'.

These are exactly what you get by substituting $x_1=a\sec\theta$, $y_1=b\tan\theta$ into the Cartesian formulas from the previous section — they're the same lines, just packaged differently.

Condition for a line to be a tangent. A natural question going the other way: given a line $y=mx+c$, for which values of $c$ (with slope $m$ fixed) does it actually touch the hyperbola, rather than missing it or cutting through both branches? Substituting $y=mx+c$ into $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ and demanding the resulting quadratic in $x$ have equal roots (the algebraic signature of tangency) gives the condition

Solving for $c$, the tangents to the hyperbola with a given slope $m$ can be written directly as

A few consequences worth noting: this only produces a real value of $c$ when $m^2 > b^2/a^2$, so a line whose slope is too shallow (in particular, a horizontal line, $m=0$) can never be tangent to the hyperbola — it always either misses the curve or cuts through one branch twice. The two vertical tangents, $x=\pm a$, have to be handled separately since they don't fit the $y=mx+c$ form at all.

Worked example — condition for tangency. For $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ ($a^2=9,b^2=4$), find the tangents with slope $m=1$. Using $c^2=a^2m^2-b^2 = 9(1)-4=5$, so $c=\pm\sqrt5$. The two tangents are $y=x+\sqrt5$ and $y=x-\sqrt5$ — a matching parallel pair, one touching each branch. …