# Intuition of the equation of a hyperbola

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.

• Wikipedia's illustrations may help too (intersection of a plane and a conic). – Raymond Manzoni Sep 14 at 10:53
• Yup, I know that the hyperbola is a conic section but the problem is that I don't know why this equation arises – helpme Sep 14 at 16:08
• 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... – Raymond Manzoni Sep 15 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$$x=\pm a\sqrt{1+\frac{y^2}{b^2}}.$$