Maximum distance from a point in hyperbola $ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes

Question

The canonical hyperbola is $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

I have already derive the equation for the distance from any point $(x_0,y_0)$ in hyperbola to the asymptote $ y=\frac{b}{a}x$ as $$ d =\frac{|bx_0-ay_0|}{c}$$ where $a^2+b^2=c^2$ and $x_0 \geq a$.

I know that the maximum distance is when $x_0=a$ but I don't know how to proof it. Can you help me? Thanks.

Answer 1

WLOG $a=c\cos u,b=c\sin u$

and any point $P$ on the hyperbola can be $(a\sec v, b\tan v)=(c\cos u\sec v,c\sin u\tan v)$

So, the distance of $P$ from $bx-ay=0$ will be $$\dfrac{|b(c\cos u\sec v)-a(c\sin u\tan v)|}{\sqrt{a^2+b^2}}$$

$$=\dfrac{|c\sin u|(c\cos u\sec v)-|c\cos u|(c\sin u\tan v)}{|c|}$$

$$=|c\sin u\cos u(\sec v-\tan v)|$$

$$=\dfrac{|c\sin2u\tan\left(\dfrac v2-\dfrac\pi4\right)|}2$$

which will be maximum

if $c\sin2u\ne0$ and

if $\tan\left(\dfrac v2-\dfrac\pi4\right)\to\infty$

which occurs if $\dfrac v2-\dfrac\pi4\to\pm\dfrac\pi2$

Answer 2

By definition distance to an asymptote goes to zero for a point P on a hyperbola in first quadrant.

The asymptotes are a pair of straight lines that include $ 2\alpha $ between them where $\tan \alpha = \frac {b}{a},$ seen in standard Wiki reference picture below.

Reckon distance $s$ from center ( intersection of asymptotes) in second quadrant. By dropping a perpendicular on the second asymptote from P we have distance $ p=s \tan 2\alpha $ which tends infinity as $s \to \infty.$

These two distances green and blue can be taken as minimum and maximum distances respectively as P tends to the asymptotic limits.

( btw their product is finite.. which matter is outside this query).

Answer 3

Note that the distance to the pair of the asymptotes is defined as the shorter of the two

$$ d =\min \left(\frac{|bx_0-ay_0|}{c} , \frac{|bx_0+ay_0|}{c}\right) $$

Let the point on the hyperbola be $x_0=a\cosh t$ and $y_0=b\sinh t$. Then

$$d = \frac{ab}c \min(|\cosh t - \sinh t|, |\cosh t + \sinh t|) =\frac{ab}c\left\{ \begin{array}{c} \cosh t - \sinh t , \>\>\>t\ge0\\ \cosh t +\sinh t,\>\>\> t<0\ \end{array} \right.$$

which is an even function of $t$ and has its maximum at the peak $t=0$, i.e. $x_0=a$.