sorry, another q (1 Viewer)

[spikestar]

x^2=4(y+1)(2y+1) how can i make it ino and inverse of a parabola

 

[Mountain.Dew]

i assume that u want to get this into the form of an 'inverse' of a parabola. i think what u mean is (by looking at the equation) a hyperbola, in the form x^2 / a^2 - y^2/b^2 = 1

so here goes:

x^2=4(y+1)(2y+1) --> expanding --> x^2 =4(2y^2 + 3y + 1)

now we have to complete the square:

2y^2 + 3y + 1 = 2(y^2 + 3/2y + 1/2)
=2(y^2 + 3/2y + [(3/4)^2] + 1/2 - (3/4)^2)
=2[(y+ 3/4)^2 -1/16]

so we get: x^2 = 8[(y+ 3/4)^2 -1/16] ==> x^2 = 8(y+ 3/4)^2 - 1/2

8(y+ 3/4)^2 - x^2 = 1/2

so 16(y+ 3/4)^2 - 2x^2 = 1 ....w00t

however, if u get the INVERSE of the original function, then it looks remarkably similar...

16(x + 3/4)^2 - 2y^2 = 1 <== looks kinda 'hyperbolish' to me

i think that this is what spikestar wanted. please correct me im wrong.