Recall that the parabola x^2=4py has
vertex at (0,0)
focus at (0,p)
Clearly your parabola has been rotated through an angle θ where tanθ = 6/2
The distance from (0,0) to (2,6) is √40=2√10.
So, in the rotated coordinate system (x',y') we have
y' = 8√10 x'^2
Now, if you haven't yet studied rotation of axes, then you must have a typo. Otherwise, if the vertex is at (0,0) then the focus must be on one of the axes.
help please!! write an equation of a parabola with vertex at the origin and the given focus
focus at (2,6)??? Please help im lost!!
3 answers
it says the focus is at (2,6) and vertex is at origin. We covered a little bit about rotation of axes
Oh, well, in that case, we can proceed. I misspoke. The way I rotated the axes, I should have said
x' = 8√10 y'^2
I'd rather have y=ax^2, so I should have said
x'^2 = 8√10 y'
or
y' = 1/(8√10) x'^2
I'll get back to you on how that works with the rotation of axes. Since the focus is at (2,6) it means that the axes have been rotated clockwise through an angle θ such that tanθ = 2/6.
So,
sinθ = -1/√10
cosθ = 3/√10
You can use those values in your rotation matrix.
x' = 8√10 y'^2
I'd rather have y=ax^2, so I should have said
x'^2 = 8√10 y'
or
y' = 1/(8√10) x'^2
I'll get back to you on how that works with the rotation of axes. Since the focus is at (2,6) it means that the axes have been rotated clockwise through an angle θ such that tanθ = 2/6.
So,
sinθ = -1/√10
cosθ = 3/√10
You can use those values in your rotation matrix.
3 answers