0

enter image description hereI have ellipse, lets say that the height is half of its width and the ellipse is parallel to x axis. then the lets say the center point is situated in the origin (0, 0) and 20 degrees from that point is lets say (4, 2). I am searching for a formula for finding the semiminor and semimajor axis (aka half of width and half of height of the ellipse)... I hope you guys can help me.

can i use this? http://www.oocities.org/web_sketches/ellipse_notes/ellipse_slope/ellipse_slope_formula.html

Chinchan Zu
  • 147

2 Answers2

2

You will need a second point that is not symmetric to your first. Reason is you need to solve for two unknowns, and each point will generate only one equation.

Once you have a second point, you simply plug x and x for both points into this equation
x^2/a^2 + y^2/b^2 = 1
and then solve for a and b from your two simultaneous equations.

Update:
If the eccentricity e is given, then the equation a = e * b will stand in for a second point.

Pieter Geerkens
  • 901
  • The statement that the ratio of axes is $2$ is enough. That gives $\frac ab$ so now we only have one unknown. – Ross Millikan Apr 22 '13 at 03:28
  • @RossMillikan: Than you; I interpreted OP's comment on the eccentricity differently. As you point out, a given value for eccentricity is equivalent to a second (non-symmetric) point. – Pieter Geerkens Apr 22 '13 at 03:37
  • hi pieter, thanks for your answer. sorry im really bad at math. I am making a game. the character is placed randomly on the screen and based on the current position of the character I should rotate it on an ellipse with the center situated on the center of screen. and the ellipse height is half of the width (based on settings, sometimes 0.4 as big).... so i can only provide the random (x,y) position and the angle of it from origin – Chinchan Zu Apr 22 '13 at 03:42
  • Ignore the angle from the origin - it is redundant (ie completely determinable from the (x,y) coordinates of the point). the ratio of width to height is the eccentricity mentioned above. Only basic Grade 9 algebra is required to solve for a and b given either two points or one point and an eccentricity value. – Pieter Geerkens Apr 22 '13 at 03:48
  • Hi man, thanks for your answer. Unfortunately these kind of problem is too complex for me. sorry for that. anyway, from the answer of Ross above, he answered that the semimajor axis is sqrt(32) = 4(sqrt(2))... i dont clearly understand howd he lead to that answer given the formula and value for square or r = 8. can you help? thanks man – Chinchan Zu Apr 22 '13 at 07:42
1

Your $20^\circ$ is not consistent with $(4,2)$ as $\arctan \frac 12 \approx 26.565^\circ$. If we accept that the ratio of the axes is $2$, the equation of the ellipse is $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$ Putting the point on the ellipse we get $r^2=8$ so we get the equation $\frac {x^2}{32}+\frac {y^2}8=1$ and the semi-major axis is $\sqrt {32}= 4\sqrt 2$, the semi-minor axis is $\sqrt 8=2\sqrt 2$

Ross Millikan
  • 374,822
  • Hi ross thanks for your answer and sorry if the image is confusing. it is just an assumption so its not 100% accurate. so suppose (4,2) is situated 26.6 degrees from the origin, how can i change your formula if the height is 0.4 as big as width? – Chinchan Zu Apr 22 '13 at 03:29
  • 1
    @ChinchanZu: The image was not confusing, you just overspecified the problem. I chose to ignore the $20^\circ$ and assumed $(4,2)$ was on the ellipse. The $4$ in my equation was the square of the ratio of the axes. So if the minor axis is $0.4=\frac 25$ of the major axis, the $4$ becomes $\frac {25}4=6.25$ – Ross Millikan Apr 22 '13 at 03:33
  • wow man! i love you. I will try your equation and let you know about the result. I am actually making a game that rotates to an ellipse and the ellipse changes based on the distance of character. Im a programmer and really suck at math. thanks man – Chinchan Zu Apr 22 '13 at 03:35
  • another question man, sorry for my stupidity... can you tell me what r stands for and why is it 8 in your first equation? – Chinchan Zu Apr 22 '13 at 03:56
  • @ChinchanZu: It is a scale factor for the size of the ellipse. At that point I know the ellipse is twice as wide as high, which is where the $4$ comes from. The standard equation for an ellipse is $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$, with $a,b$ the semi-axes. You told me $\frac ab=2$, so $a^2=4b^2$ (and I changed $b$ to $r$). I haven't made use of the fact that $(4,2)$ is on the ellipse yet. Then I plug $(4,2)$ in and get an equation for $r$ which simplifies to $r^2=8$. – Ross Millikan Apr 22 '13 at 04:00
  • Thank you man for the explanation. I tried figuring out myself the equation and also understood how you got (r)2 = 8. but what i dont understand is how did you substituted the (r)2 and the equation to get the value of the semi-minor axis which is the squareroot of 32? thanks again man and sorry if i sound really stupid at math – Chinchan Zu Apr 22 '13 at 07:49
  • Given $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$, if we substitute in $x=4,y=2$ we have $\frac 4{r^2}+\frac 4{r^2}=1$ or $\frac 8{r^2}=1$ so $r=\sqrt 8$ The last two come from the relation $\sqrt {a^2b}=a \sqrt b$ You can pull things out of the square root sign if you take the square root. – Ross Millikan Apr 22 '13 at 13:24
  • got it man. this finally solved the problem! thanks a bunch! – Chinchan Zu Apr 23 '13 at 02:18