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@jim_thompson5910 please verify
The parabola will open to the left side, so it will have a y^2 term (instead of an x^2 term)
not use to those so the vertex is (0,0)
I'm using the ideas shown here http://www.mathwords.com/f/focus_parabola.htm 4p(x-h) = (y-k)^2 4*8(x-0) = (y-0)^2 32x = y^2 x = (1/32)*y^2 x = -(1/32)*y^2 ... the coefficient is made to be negative so the parabola opens to the left (instead of right) |dw:1433217556903:dw|
so is the equation I have wrong ?
anything with x^2, and not y^2, will open upward or downward
I actually have the same diagram see above (not as neat )
I see it, but again, you'll have y^2 instead of x^2
you are saying since I have the directrix vertical instead of horizontal then the value from f to v is the y component?
because the directrix is vertical, this means that the parabola will open to the left or right https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A7980ac3c31917d6d9238e3fe5ee79be178774387290a21eba7ba82d2%2BIMAGE%2BIMAGE.1
got that not the problem what I am missing is how to write the equation
http://www.mathwords.com/f/focus_parabola.htm you use 4p(x-h) = (y-k)^2 since the parabola opens to the left, p < 0 based on that previous image. So p = -8 (h,k) = (0,0) is the vertex
@jim_thompson5910 thank you so much !!! I will look at the links guess better try to learn that version of the equation thank you
So what would the finally equation be?
follow what Jim had above. I am used to the other format and in addition I had my x and y switched since it opened left, sorry
It depends on what format they want it in, but you could go with something in the form x = ay^2 + by + c