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Given any second degree equation in $x$ and $y$,

$ax^2+by^2+2gx+2fy+2hxy+c=0$

is it possible to find out the centre and/or the axis of the conic section it represents?

What information can I gather simply by looking at the curve?

For example one thing my teacher taught me was that if $h=0$ the axis of the curve(be it parabola,hyperbola,ellipse etc) is parallel to the coordinate axes.

I am looking for some other information that the equation can offer.

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  • $\begingroup$ You could start with taking a look at this page. $\endgroup$ – StackTD Apr 26 '16 at 9:16
  • $\begingroup$ :| Everything on that page seems so complex.I am still in high school so I didn't understand all that.I would be grateful if you could simplify those statements on the page.I do have a working knowledge of partial derivatives and matrices/determinants $\endgroup$ – Karan Singh Apr 26 '16 at 9:18
  • $\begingroup$ I don't have the time to go into detail. If $h=0$, it is easy to transform the general equation into one of the standard equations of the different conic sections, the type of conic and its properties (center, axes etc) follow immediately then. A useful trick to turn the equation into a standard form is completing the square. If $h \ne 0$, you'll need to perform a rotation to get rid of this mixed term. $\endgroup$ – StackTD Apr 26 '16 at 9:23
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For obtaining the center (if there is one: if it is a parabola, there is no center), compute the partial derivatives (ask your teacher if necessary) of $f$ with respect to $x$, then to $y$, and write that they are zero: you will get a system of 2 linear equations with two unknowns, the solution of which is the coordinates of the center.

Let us take the example of $f(x,y)=4x^2 + y^2 + 6x y - 4 x=0$ (plotted below: this is a hyperbola). The partial derivatives are:

$$\begin{cases}\frac{\partial f}{\partial x}&=&-4+8x+6y\\\frac{\partial f}{\partial y}&=&6x+2y\end{cases}$$

Equating these expressions to zero, we obtain $x=-2/5, y=6/5$ that can be verified to be the coordinates of the center of the hyperbola.

enter image description here

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  • $\begingroup$ Thanks.Can I find the axis of the curve? $\endgroup$ – Karan Singh Apr 26 '16 at 13:41

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