Write the equation of the conic section x^2+6^2=16 after a rotation of 45 degrees about the origin.
I don't know where to begin on this problem. Could you help? Thanks.
Of course! To begin solving this problem, we need to understand how a rotation of 45 degrees about the origin affects the equation of a conic section.
Firstly, we need to know the general equation of a conic section centered at the origin, which is given as:
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
When a conic section is rotated about the origin, the xy term (Bxy) disappears. This is because rotation only affects the x and y coordinates, not their product (xy). Therefore, the equation after rotation simplifies to:
Ax^2 + Cy^2 + Dx + Ey + F = 0
Now let's apply this to your specific equation x^2 + 6^2 = 16 after a rotation of 45 degrees.
Step 1: Rewrite the equation x^2 + 6^2 = 16 as Ax^2 + Cy^2 + Dx + Ey + F = 0 format.
- x^2 + 6^2 = 16
- x^2 + 36 = 16
- x^2 = -20
Step 2: Separate the x and y terms.
- Move the x^2 term to the left side:
x^2 + 20 = 0
- There are no y terms or xy terms in the original equation, so D, E, and F are all zero.
The equation is now in the form Ax^2 + Cy^2 + Dx + Ey + F = 0. Since the equation is rotating about the origin, the xy term disappears. Thus, we have:
A = 1, C = 0, D = E = F = 0
Finally, the equation after a rotation of 45 degrees becomes:
x^2 + 0y^2 + 0x + 0y + 0 = 0
or simply:
x^2 = 0
Therefore, the equation of the conic section x^2 + 6^2 = 16 after a rotation of 45 degrees about the origin is x^2 = 0.