# Smooth curve orthogonal to all hyperbolae $xy = a$ at points of intersection.

Suppose a smooth, connected curve $C$ in $R^2$ is orthogonal to all hyperbolae $xy = a$ whenever they coincide. I'd like to find the point(s) of intersection of $C$ with the hyperbola $xy = 16$ given that $C$ contains the point (1,1).

I figure that I should write the curve $C$ as some general parametrization $C(t) = (p(t),q(t))$ and use the fact that, if $C(t)$ intersects the hyperbola $H_a(t) = (t,a/t)$ at $t = t_0$, then $$C'(t_0) \cdot H'_a(t_0) = tp(t_0) + (a/t_0)q(t_0) = 0.$$

Also, since both $C(t)$ and $H_1(t)$ contain the point (1,1), this gives me some specific information about where $C(t)$ is orthogonal to the particular hyperbola $H_1(t)$.

This question is from a practice exam and does not seem very advanced, but I keep getting stuck. If someone could give me a nudge in the right direction, I'd be grateful. Thanks.

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The orthogonal relation should be applied to tangent vectors $C'(t)$ and $H'_a(t)$. –  enzotib Aug 2 '12 at 17:29
Ah, oops. Thanks for the catch, enzotib. –  user36387 Aug 2 '12 at 17:37
Sorry, I saw after I posted a full solution that you asked for a nudge. I deleted it. Here's a nudge: Forget about calculus and think what simple curve through $(1,1)$ might be orthogonal to all these hyperbolas. Note that they have mirror symmetry about the diagonal on which $(1,1)$ lies. –  joriki Aug 2 '12 at 17:45
Well, it seems the time for nudging has passed -- I undeleted my answer. –  joriki Aug 2 '12 at 18:16

Nudge: Look at some of the graphs of $xy=a$ for various $a$. Do you notice a symmetry about some sort of axis? Algebraically, $x$ and $y$ can be switched in any point on a given such hyperbola, and we end up with another point on the very same curve. There is one curve in which all its individual points stay fixed under this 'coordinate-switching' action. The point $(1,1)$ is on it.

Where does it intersect $xy=16$?

General solution: For $(x,y)$ on the parametrized family of curves $xy=a$ we differentiate to obtain

$$y \dot{x}+x\dot{y}=0=(y,x)\cdot(\dot{x},\dot{y}).$$

(Each curve is itself parametrized by $t$ in some arbitrary fashion; the resulting implicit differential equation however does not depend on $a$ i.e. on which curve in the family we're talking about.)

Geometrically, this says $(y,x)\perp(\dot{x},\dot{y})$ (they are perpendicular vectors). Suppose we want $(u,v)$ to make up a curve perpendicular to these curves. Then we want $(v,u)\|(\dot u,\dot v)$ - that is we want them to be parallel rather than perpendicular. In two dimensions, each vector has a unique perpendicular vector up to magnitude, and two vectors being parallel is equivalent to each being perpendicular to the other's perpendicular vector. Thus $(v,u)\|(\dot u,\dot v)\iff (-u,v)\perp(\dot u,\dot v)$ (we applied a $90^\circ$ rotation matrix to obtain $(-u,v)$ as a perpendicular vector to $(v,u)$). In dot product notation, this is

$$0=(-u,v)\cdot(\dot u,\dot v)=-u\dot u+v\dot v.$$

Implicitly integrating both sides yields $v^2-u^2=C$, where $C$ is an arbitrary constant.

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A nudge and a general solution: what more could I ask for? Thanks a lot, Anon. –  user36387 Aug 2 '12 at 18:15

Sounds tricky to me, I get a solution with differential equations:

Assume we parameterize $C$ as $C(x)=(x,y(x))$ and $H_a(x) = (x,a/x)$.

Take any point $P = (x,y(x))$ on $C$, then $P$ is also on the particular hyperbola $H_a$ for $a=xy(x)$. Therefore, the tangent vectors of $C$ and $H_a$ are orthogonal. We've got

$$H_a'(x) = (1, -\frac{a}{x^2})$$ $$C'(x) = (1, y'(x))$$

$$1 - \frac{a\cdot y'(x)}{x^2}=0$$ and since we know $a$ $$1 - \frac{y(x)y'(x)}{x}=0$$

$$\Leftrightarrow y'(x)=\frac{x}{y(x)}.$$

The general solution to this is

$$y(x) = \pm\sqrt{x^2+C}.$$

Because $y(1)=1$ we get $C=0$, i.e. in the positive quadrat it's $C(x)=(x,x)$. The intersection with $xy=16$ therefore is $(4,4)$.

Is there a simpler approach?

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When you substitute $a$, you need an $y(x)$ instead of $x$ in the numerator. –  enzotib Aug 2 '12 at 18:02
Thanks for the quick response, Dario. This approach seems simple enough to me. There is, however, a small typo I believe: I think you wrote $x$ instead of $y(x)$ in the numerator of a fraction you typed. –  user36387 Aug 2 '12 at 18:14
Thanks, fixed the typo. –  Dario Aug 3 '12 at 8:47

The curve $y_a(x)=a/x$ has derivative $y_a'(x)=-a/x^2$. If a curve $f(x)$ coincides with $y_a(x)$, we have $f(x)=a/x$, so $a=xf(x)$ and $y_a'(x)=-xf(x)/x^2=-f(x)/x$. For the curves to be orthogonal we must have $f'(x)*y_a'(x)=-1$, and thus $f'(x)f(x)=x$. The general solution of this ordinary differential equation is $f^2(x)=x^2+C$, or $f(x)=\pm\sqrt{x^2+C}$.

If the curve contains the point $(1,1)$, we have $C=0$, and in fact we could have guessed the solution $y=x$ without doing any calculus. The intersection with $xy=16$ is at $(4,4)$.

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