This is a question about hyperbolas

conic sections

The question goes like:

A general hyperbola has its axes at right angle and the asymptotes have reflection symmetry about these axes. However, the axes may be rotated, unlike the standard form of a hyperbola. The general equation of a hyperbola can be written as $ax^2+bxy+cy^2+dx+ey+f=0.$

The general form of a hyperbola passing through the point $(x_0,y_0)$ can be derived from the asymptotes using the hyperbola equations. $(A_1x+B_1y+C_1)(A_2x+B_2y+C_2)=(A_1x_0+B_1y_0+C_1)(A_2x_0+B_2y_0+C_2)$ where $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ are the two asymptotes.

Find the general equation of a hyperbola which passes through the point $(1,1)$ has an asymptote $y=0$ and has an axis of symmetry $y=2x+2$.

Any help will be really appreciated- thank you!

Best Answer

From the reflective property of the asymptotes with respect to the axes of the hyperbola, we can find the equation of the other asymptote.

The two asymptotes and the axes of symmetry all meet in a single point. And this point is the center of symmetry of the hyperbola.

We're given that $y=0$ is an asymptote, and that $y = 2 x + 2 $ is an axis. These two lines intersect at $ x = -1, y = 0 $, so the center of the hyperbola is the point $(-1, 0)$.

Now we want to refect the line $y = 0$ about the line $ y = 2 x + 2 $, a small sketch will reveal that the reflected line will have a slope of $ \tan( 2 \tan^{-1}(2) ) = \dfrac{2(2)} {(1 - 4} = - \dfrac{4}{3} $

Hence the reflected line (the other asymptote) equation is $ y = -\dfrac{4}{3} x + b $

Since the hyperbola center $(-1,0)$ is on this reflected line, then $ b = -\dfrac{4}{3} $. Hence, the equation of the other asymptote is $ y = - \dfrac{4}{3} (x + 1) $

From here, we just apply the formula given in the question to find the equation of the hyperbola, using the point $(x_0, y_0) = (1, 1) $

$ ( y ) (\dfrac{4}{3} x + y + \dfrac{4}{3} ) = (1) (\dfrac{4}{3}(1) + (1) + \dfrac{4}{3} ) $

And this simplifies to

$ ( y ) (\dfrac{4}{3} x + y + \dfrac{4}{3} ) = \dfrac{11}{3}$

Multiplying the brackets on the left side and multiplying through by $(3)$, results in the final form of the equation of the hyperbola

$ 4 xy + 3 y^2 + 4 y - 11 = 0 $

Related Solutions

The general equation of a rectangular hyperbola whose axes pass through origin

The hyperbola given by the equation $xy=c^2>0$ is your prime example of a rectangular hyperbola. Let's rotate it by some arbitrary angle $\phi$! This rotation maps points $(x,y)$ to new points $(x',y')$, whereby $$(x,y)^\top=\left[\matrix{\cos\phi&-\sin\phi\cr\sin\phi&\cos\phi\cr}\right](x',y')^\top\ .$$ This means that the coordinates of the preimage $(x,y)$ are obtained from the coordinates $(x',y')$ of the image by $$x=\cos\phi\> x'-\sin\phi\> y',\qquad y=\sin\phi\> x'+\cos\phi\> y'\ .$$ If $(x,y)$ is lying on the model hyperbola then $xy=c^2$. This implies that the coordinates of the image point $(x',y')$ have to satisfy $$(\cos\phi\> x'-\sin\phi\> y')(\sin\phi\> x'+\cos\phi\> y')=c^2\ .$$ This can be rewritten as $$\cos\phi\sin\phi(x'^2-y'^2)+(\cos^2\phi-\sin^2\phi)x'y'=c^2\ .$$ Leaving away the primes here gives the result of the question.
$$\Rightarrow (x^2-y^2)\sin\theta+2xy\cos\theta=2c^2$$

Express by a single equation a region delimited by confocal ellipse and hyperbola

Such a curve having the shape of a double-sided axe (https://en.wikipedia.org/wiki/Labrys) with mixed elliptical/hyperbolic boundaries can be given by the general equation

$$x^2+k|y^2-1|=k^2 \tag{1}$$

with a "behavorial change" when variable $y$ crosses values $y=1$ or $y=-1$ as can be seen on the following graphics for different values of $k$:

Fig. 1: Curves with equation (1) for different values of $k$, with $k=0.25$ for the most internal one (in two parts) and $k=2$ for the most external, with steps $0.25$. For our question, only values $k>1$ (blue curves) make sense.

It remains now to tilt the "axe" using the following coordinates change in (1):

$$\begin{cases}x&=& \ \ \ (\cos \theta) X + (\sin \theta) Y \\ y&=&-(\sin \theta) X + (\cos \theta) Y \end{cases}$$

(It looks like a pun: a change of axes inducing a different axe...)

Explanations:

I have obtained (1) by imposing common foci to the ellipse and the hyperbola. How can it be done ? If $f$ is the common distance from the origin to the foci, it is known that $f^2=a^2-b^2$ for an ellipse and $f^2=a^2+b^2$ for a hyperbola with canonical equations $\dfrac{x^2}{a^2}\pm\dfrac{y^2}{b^2}=1$ ; one obtains in particular $f=\sqrt{k^2-1}$.
As a consequence, the elliptical arcs are orthogonal to the hyperbolic arcs. This wasn't necessary (see the solution by @David K).
Case $k=1$ is particular as can be seen on Fig. 1 : the elliptical arcs become circular arcs.

Edit: Here is the Matlab program I have written for the generation of the figure:

   clear all;close all;hold on;axis equal
   for k=0.25:0.25:2;
      c='r';
      if k>1;c='b';end;
      e=ezplot([num2str(k),'*abs(y^2-1)+x^2-',num2str(k^2)]);
      get(e);set(e,'linecolor',c);
   end;
   u=2.5;plot([-u,u],[0,0],'k','linewidth',1.5);
   v=2;plot([0,0],[-v,v],'k','linewidth',1.5);

Best Answer

Related Solutions

The general equation of a rectangular hyperbola whose axes pass through origin

Express by a single equation a region delimited by confocal ellipse and hyperbola

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