Question #321647

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?

1
Expert's answer
2022-06-30T12:31:13-0400

4(x2+2y)=9    y=9812x24(x^2+2y)=9\implies y=\frac{9}{8}-\frac{1}{2}x^2 , hence the conic is a parabola.

Let (x1,y1)(x_1,y_1)  be the new coordinates of point (x,y)(x,y)  after the shift of the origin and let (x2,y2)(x_2,y_2)  be the new coordinates of (x1,y1)(x_1,y_1)  after the rotation, then

x=x1+1x=x_1+1

y=y11y=y_1-1

x1=x2cos(π/3)y2sin(π/3)=12x232y2x_1=x_2cos(\pi/3)-y_2sin(\pi/3)=\frac{1}{2}x_2-\frac{\sqrt{3}}{2}y_2

y1=x2sin(π/3)+y2cos(π/3)=32x2+12y2y_1=x_2sin(\pi/3)+y_2cos(\pi/3)=\frac{\sqrt{3}}{2}x_2+\frac{1}{2}y_2

hence

x=12x232y2+1x=\frac{1}{2}x_2-\frac{\sqrt{3}}{2}y_2+1

y=32x2+12y21y=\frac{\sqrt{3}}{2}x_2+\frac{1}{2}y_2-1

substituting the last two equations into the original equation:

4((12x232y2+1)2+2(32x2+12y21))=9,4((\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,

(x23y2+2)2+43x2+4y28=9,(x_2-\sqrt{3}y_2+2)^2+4\sqrt{3}x_2+4y_2-8=9,

x22+3y22+423x2y2+4x243y2+43x2+4y28=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,

x22+3y2223x2y2+4(1+3)x2+4(13)y2=13.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.


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