Find the equation in normal Cartesian Coordinates for the following Conic section

Sep 2012
1,061
434
Washington DC USA
The trick is to first write down the ellipse's equation *prior to* that 45 degree rotation. Only then are you ready to see what the equation is after you've rotated it.

So, pretending for a moment that there's no 45 degree rotation there, can you write down the equation of the ellipse?
 
Sep 2012
1,061
434
Washington DC USA
A quick head calculation (I could've made a mistake too) tells me yes, you're equation is correct, because the 4 points you know all satisfy that equation for an ellipse.

(Note: \(\displaystyle B^2 - 4AC = (-10)^2 - 4(13)(13) = 100 - 4(13)^2 < 0\), so that is the equation for an ellipse.)

If you got it by "plugging in", that's correct, but I assume you're expected to know how to do it by rotations.

Did you get it by rotating \(\displaystyle \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1\) by 45 degrees? Or by plugging the 4 known points into \(\displaystyle Ax^2 + Bxy + Cy^2 = K\)?
 
Last edited:
Jul 2015
45
0
Boston, MA
yes by rotating x^2/3^2 + y^2/2^2 = 1 by 45 degrees