0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Wikipedia's illustrations may help too (intersection of a plane and a conic). $\endgroup$ – Raymond Manzoni Sep 14 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 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 at 7:57
1
$\begingroup$

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$$x=\pm a\sqrt{1+\frac{y^2}{b^2}}.$$

$\endgroup$
  • $\begingroup$ 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 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.