Answer to Question #321647 in Analytic Geometry for Rit.

Question

Consider the conic represented by 4(x²+2y)=9. Now shift the origin to (1,-1) and then rotate the axes through π/3 What is the resultant equation? What geometrical object does it represent?

Answer 1

$4(x^2+2y)=9\implies y=\frac{9}{8}-\frac{1}{2}x^2$ , hence the conic is a parabola.

Let $(x_1,y_1)$  be the new coordinates of point $(x,y)$  after the shift of the origin and let $(x_2,y_2)$  be the new coordinates of $(x_1,y_1)$  after the rotation, then

$x=x_1+1$

$y=y_1-1$

$x_1=x_2cos(\pi/3)-y_2sin(\pi/3)=\frac{1}{2}x_2-\frac{\sqrt{3}}{2}y_2$

$y_1=x_2sin(\pi/3)+y_2cos(\pi/3)=\frac{\sqrt{3}}{2}x_2+\frac{1}{2}y_2$

hence

$x=\frac{1}{2}x_2-\frac{\sqrt{3}}{2}y_2+1$

$y=\frac{\sqrt{3}}{2}x_2+\frac{1}{2}y_2-1$

substituting the last two equations into the original equation:

$4((\frac{1}{2}x_2-\frac{\sqrt{3}}{2}y_2+1)^2+2(\frac{\sqrt{3}}{2}x_2+\frac{1}{2}y_2-1))=9,$

$(x_2-\sqrt{3}y_2+2)^2+4\sqrt{3}x_2+4y_2-8=9,$

$x_2^2+3y_2^2+4-2\sqrt{3}x_2y_2+4x_2-4\sqrt{3}y_2+\\ 4\sqrt{3}x_2+4y_2-8=9,$

$x_2^2+3y_2^2-2\sqrt{3}x_2y_2+4(1+\sqrt{3})x_2+\\4(1-\sqrt{3})y_2=13.$