I 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
On
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.
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 $\sqrt 8=2\sqrt 2$