Writing equations for parabolas

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Writing equations for parabolas

Mathematics
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Write eq in standard form of the parabola that satisfies f= (4,3) directix x=-1
so vertex would be = (1.5,3) ?

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@jim_thompson5910 haha sorry for bugging you all day!
|dw:1433483905158:dw|
A faster way to do it is realize that we can pick a point on the directrix say d = (-1,3) and the f = (4,3) and then find the middle distance from -1 to 4 which is 1.5 so the v = (1.5,3)
I just need to write the equation now. Which i'm not sure how to do D:
mmm vertex looks good :)
it's horizontal so: 4p(x-h)=(y-k)^2 4p(x-1.5)=(y-3)^2 ?
how do I find p?
oh! p is the distance is from vertex to foci?
distance between focus and directrix was 5, ya? p is half of that. the value you used to find the vertex. yes.
so p = 2.5 yaya
4(2.5)(x-1.5)=(y-3)^2
how would I write that more attractively :P
10(x-1.5)=(y-3)^2
is it ok to write it as y^2=10x-18 ?
or should I write it as y= +/- sqroot 10x+-18 ?
\[(x+1)^2=(x-4)2+(y-3)^2\] \[x^2+2x+1=x^2-8x+16+(y-3)^2\] \[x^2+2x+1-x^2+8x-16=(y-3)^2\] \[(y-3)^2=10x-15\]
@surjithayer we forgot the shift upwards. o.0
shift upwards? what? 0_o
That's what the y-3 is, ya?
yeah
the focal point is (4,3) so we know that it's not on the x = 0 line
but when I put the equation into a graphing thing it puts it on the x line
it does? 0_o https://www.desmos.com/calculator/wyivsbib3n looks ok to me.
hrrm I simplified it further and it was coming out wrong. Probably my calculations were wrong. Ok, all good. Thanks :)

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