Hyperbola equation not parallel to axis

Given the foci and vertex, I want to know how to get the equation of a hyperbola whose axes are not parallel to x or y axis. All materials I have read only discuss when axes are parallel to axis.

• Have you done linear algebra, specifically operations of rotation and scaling in the plane? Geometrically what you're looking for is just taking $x^2 - y^2 = 1$ and rotating and rescaling it by replacing (x,y) by the image of (x,y) after such operations. Like (x,y) -> (x + y,x-y) is a rotation and rescaling and the graph of $(x + y)^2 - (x - y)^2 = 1$ is a rotated hyperbola. – Squid Feb 14 '17 at 8:41
• The method is going to depend on the description of the hyperbola that you’re starting with. What do you have to work with? – amd Feb 14 '17 at 10:45
• @amd, I have foci and vertex. I want to get the hyperbola equation. The hyperbola is not parallel to x or y axis and the center is not in the origin – a hidayat Feb 14 '17 at 12:24
• As this is important I edited the question, hope ok. – Narasimham Feb 16 '17 at 21:25

In your comments to the question, you say that you are given the two foci (we’ll call them $F_1$ and $F_2$) and a vertex $V$. There are several ways to proceed with these data.

A direct approach is to use the definition of a hyperbola as the set of points with a constant difference of distances from two fixed points (the foci). Since you have a point on the hyperbola, you know what this difference $d$ must be: it’s the absolute value of $\|V-F_1\|-\|V-F_2\|$. So, an equation of the hyperbola is $(\|P-F_1\|-\|P-F_2\|)^2=(\|V-F_1\|-\|V-F_2\|)^2$. You’ll need to do some work to eliminate the square roots after you substitute coordinates for all of the points.

Another approach is to use the focus-directrix form of the equation, in which the ratio of distances to a focus and a line called the directrix is a constant called the eccentricity. The eccentricity is easily computed from the given data. The center of the hyperbola is the midpoint of the foci, from which we get $$e=\frac c a = {\|F_2-F_1\|/2\over \|V-(F_1+F_2)/2\|}$$ which you can simplify as necessary. With the eccentricity in hand, you can find the directrix and set up the equation of the hyperbola. This is a bit more work than the first approach, but has the advantage that the resulting square roots are easier to eliminate.

For a hyperbola, the eccentricity $e=\sec\theta$, where $\theta$ is the angle that the asymptotes make with the transverse axis. This suggests yet another approach: If the equations of the asymptotes are $ax+by+c=0$ and $Ax+By+C=0$, then an equation of the hyperbola can be written as $(ax+by+c)(Ax+By+C)=d$ for some constant $d$. Given the eccentricity and the foci, you can find the equations of the two asymptotes and adjust $d$ so that the given vertex satisfies the equation.

However, once you’ve gone as far as computing the eccentricity $e$ and the semi-major axis length $a$, the most straightforward way to go is to compute $b$ and write down the equation ${x^2\over a^2}-{y^2\over b^2}=1$ of the hyperbola in standard position and then rotate and translate it. This is quite easy to do in (homogeneous) matrix form. The matrix of the equation in standard position is simply $$M=\pmatrix{\frac1{a^2}&0&0\\0&-\frac1{b^2}&0\\0&0&-1},$$ i.e., the equation is $[x,y,1]M[x,y,1]^T=0$. The required center is just the midpoint of the two foci, so the necessary translation is easy to compute. The rotation matrix is also easy to construct if you remember that the columns of a transformation matrix are the images of the basis vectors, so the first column will be a unit vector in the direction of the hyperbola’s transverse axis, and the second column is just the first rotated by 90°, which can be found by swapping components and negating one of them.

$$\frac{(x - h)^2}{a^2} + \frac{(x - h) (y - k)}{c^2} + \frac{(y - k)^2}{b^2} = 1$$

The hyperbola above has its center at $(h,k)$

Check how $c$ influences the tilt in relation to $(a,b)$