The standard form of a circle is:
${x^2} + {y^2}={r^2}$
Which shows that the circle is a locus of points with the same distance from the focus (centre).

The standard form of an ellipse is:
$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$
Which shows that the ellipse is a circle stretched horizontally by a factor of a, and stretched vertically by a factor of b.

The standard form of an hyperbola is:
$\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$
Which I fail to understand intuitively.

Why does swapping the sign from + to - turn an ellipse into a very different-looking hyperbola? Is there a short intuitive reason as to why?

I've been searching for answers, and came across a thread that mentioned that for an ellipse, $y\to iy$ will cause it to become a hyperbola.
It was mentioned that "the hyperbola is the family of curves of orthogonal trajectory to an ellipse".

Can someone simplify this or explain it? I know that multiplying by $i$ will cause a 90 degree rotation anticlockwise on a complex plane. Since an ellipse is on an xy-plane, does that mean that you can add an imaginary z-axis and the ellipse can somehow become a hyperbola? This is very difficult to visualise.

  • $\begingroup$ Wikipedia's illustrations may help too (intersection of a plane and a conic). $\endgroup$ – Raymond Manzoni Sep 14 '19 at 10:53
  • $\begingroup$ Yup, I know that the hyperbola is a conic section but the problem is that I don't know why this equation arises $\endgroup$ – helpme Sep 14 '19 at 16:08
  • $\begingroup$ The equation of the illustrated conic is $\;x^2+y^2= (a z)^2\;$ (horizontal circle growing with $z$ at a constant 'speed' $\,a\,$ supposed $1$ to simplify the equation to : $\;x^2+y^2= z^2\;$). If you consider $z=R$ constant then you get a circle $\;x^2+y^2= R^2\ (2)$ while if you consider $x=R$ constant you get a hyperbola $\;R^2=z^2-y^2=(z-y)(z+y)\ (3)$. If $\,z-x=R\,$ constant you get $\,y^2=(2x+R)\,R\;$ a parabola $(1)$. This is a start even if not what you wished... $\endgroup$ – Raymond Manzoni Sep 15 '19 at 7:57

Consider the expression$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$If it holds for a pair $(x,y)$, then $\lvert y\rvert$ cannot be very large. In fact, we can't have $\lvert y\rvert>\lvert b\rvert$, because then $\frac{x^2}{a^2}<0$, which is impossible.

But if you consider the expession$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$then there os no restricton on $y$. No matter how large it is, you can always take$$x=\pm a\sqrt{1+\frac{y^2}{b^2}}.$$

  • $\begingroup$ It is possible to get an understanding of how the x and y values change throughout the conic section (such as an ellipse and a hyperbola), but the main concern is how the equation of the hyperbola arises in the first place (before even seeing how it changes throughout as the x and y values change) $\endgroup$ – helpme Sep 14 '19 at 16:10

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