Suppose that a quadratic equation in two variables is put into standard form using the following changes of variables: a rotation x = Px' where P is the matrix representing (anticlockwise) rotation by 5*pi/4, and a translation given by x'' = x'+2, y'' = y'-3.

In the new variables x'', y'' it is a parabola with vertex at (0,0) with the x" axis as the axis of symmetry,and which also passed through the point (x",y")=(3,1).

Find the original quadratic equation in terms of x and y. Anyone could give me a hand with this question ? Thanks.

Question

Suppose that a quadratic equation in two variables is put into standard form using the following changes of variables: a rotation x = Px' where P is the matrix representing (anticlockwise) rotation by 5*pi/4, and a translation given by x'' = x'+2, y'' = y'-3.

In the new variables x'', y'' it is a parabola with vertex at (0,0) with the x" axis as the axis of symmetry,and which also passed through the point (x",y")=(3,1).

Find the original quadratic equation in terms of x and y. Anyone could give me a hand with this question ? Thanks.

dan815 · Answer

okk

dan815 · Answer

wat are u confused about

anonymous · Answer

Is the rotation given by:
$$x=Px' \text{ or  } x'=Px \text{ ?}$$

anonymous · Answer

I have it sketched out but I feel like you're missing one thing to finish it. I have the x and y coordinates in terms of the x'' and y'' coordinates so given (x'',y'') you can find x and y but from there I'm not quite sure.

anonymous · Answer

I get:
$$x = \frac{1}{\sqrt{2}} (-x'' + y''+5) \text{ and } y = \frac{1}{\sqrt2}(-x''-y''-1)$$

anonymous · Answer

Ah! From there you plug in the vertex point and the axis of symmetry point and since the vertex is ON the symmetry axis you can find the equation of the line by finding the slope and using one of the points! This will tell you the orientation of the parabola in x-y space.