Asymptotes of a Hyperbola

What is an asymptote? Consider the two lines through the origin, $y=\pm\dfrac{b}{a}x$ (equivalently $\dfrac{x}{a}=\pm\dfrac{y}{b}$). Now imagine a point $P$ moving along the hyperbola, further and further from the centre. As $P$ moves out, the perpendicular distance from $P$ to one of these two lines keeps shrinking — getting closer and closer to zero — but, crucially, it never actually reaches zero; the branch of the hyperbola gets arbitrarily close to the line without ever touching or crossing it. Such a line, approached but never reached, is called an asymptote of the hyperbola.

Every standard hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has exactly two asymptotes, $y=\dfrac{b}{a}x$ and $y=-\dfrac{b}{a}x$, both passing through the centre and both shared with its conjugate hyperbola $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$ (which is why the conjugate hyperbola's two branches nestle into the "other" two regions carved out by the same pair of asymptote lines).

Worked Example 1 — tangent at a point in a given quadrant. Hyperbola $2x^2-3y^2=5$; a point in the third quadrant has abscissa $-2$: $2(4)-3y^2=5\Rightarrow 3=3y^2\Rightarrow y=\pm1$; third quadrant needs both coordinates negative, so $P=(-2,-1)$. Writing the hyperbola as $\dfrac{x^2}{5/2}-\dfrac{y^2}{5/3}=1$ ($a^2=5/2,b^2=5/3$), the tangent (using the direct-substitution rule for $Ax^2-By^2=C$, tangent $=Axx_1-Byy_1=C$): $2x(-2)-3y(-1)=5\Rightarrow-4x+3y=5\Rightarrow4x-3y+5=0$.

Worked Example 2 — verifying a tangent and finding the point of contact. Line $4x-3y=16$; hyperbola $16x^2-25y^2=400$, i.e. $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$ ($a^2=25,b^2=16$). Line: $y=\dfrac43x-\dfrac{16}{3}$, so $m=\dfrac43,c=-\dfrac{16}{3}$. Check: $a^2m^2-b^2=25\left(\dfrac{16}{9}\right)-16=\dfrac{400}{9}-\dfrac{144}{9}=\dfrac{256}{9}=\left(\dfrac{16}{3}\right)^2=c^2$ — confirmed tangent. Point of contact $=\left(-\dfrac{a^2m}{c},-\dfrac{b^2}{c}\right)=\left(-\dfrac{25(4/3)}{-16/3},-\dfrac{16}{-16/3}\right)=\left(\dfrac{25}{4},3\right)$. …