Question

OK, so I have the equation for a parabola with focus at sqrt(2) and directrix y=0

y^2=2sqrt(2)*(x-sqrt(2)/2)

So how can I rotate that equation 45 degrees to find the equation of a parabola with focus (1,1) and directrix y=-x ?

Answer 1

to rotate it -45 degrees, in order to match the slope of y=-x

we add cos(-45) to each x value and the value of sin(-45) to each y value i believe

Answer 2

thats the thought anyways, id have to dbl chk it tho

Answer 3

i cant remember how to do this without linear algebra, lmao

Answer 4

actually, the way you wrote it is good, but you have it already turned on its side; so add cos(45) to x and sin(45) to y ... still havent double checked it tho

Answer 5

OK, I need to do this algebraically. I can't use trig yet.

Answer 6

that didnt do it ... well, sqrt(2)/2 is the values, but so far that just shifts the graph and not rotate it, let me see what I am forgetting

Answer 7

to convert rectangular to polar x=cos(t) and y=sin(t) and r=sqrt(x^2 + y^2)

Answer 8

we want to keep the same r value but move the points to the right

Answer 9

and by right i mean left :)

Answer 10

http://www.mecca.org/~halfacre/MATH/rotation.htm

Answer 11

ugh .... i cant make any sense of it :/ if you never use it, you lose it