Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.

Question

anonymous · Answer

no one likes these conic section problems, do they?

anonymous · Answer

not really :p

anonymous · Answer

first lets see what it looks like

anonymous · Answer

ok isn't it pointing up?

anonymous · Answer

|dw:1396487873299:dw|

anonymous · Answer

oh never mind that then

anonymous · Answer

no i don't think so
the directrix is a vertical line, and the focus is 8 units to the left
can you guess which way it opens now?

anonymous · Answer

it opens to the left right?

anonymous · Answer

yes. which means the $y$ term is squared

anonymous · Answer

ok then i have to use the parabola equation right?

anonymous · Answer

right 
now you need to find two things, first the vertex

anonymous · Answer

the vertex is half way between the focus $(-4,0)$ and the directrix $x=4$

anonymous · Answer

is it clear what that is?

anonymous · Answer

so that would be (0,0)

anonymous · Answer

yes, which makes life much easier, because instead of being $4p(x-h)=(y-k)^2$ it is just 
$$4px=y^2$$

anonymous · Answer

now we need $p$ which is the distance between the vertex and the focus, or the distance between the vertex and directrix (since it is halfway, they are the same thing)

anonymous · Answer

so 4?

anonymous · Answer

since it opens to the left, we use $-4$

anonymous · Answer

well the distance is 4 you are right, but as it opens to the left we want $-4$ now just plug that in to 
$$4px=y^2$$ and you will have it

anonymous · Answer

want to check it to see if it is right?

anonymous · Answer

i plug the -4 into  the p right?

anonymous · Answer

right you should get $$-16x=y^2$$ and here is the check that it is right scroll down and you will see the vertex and the directrix http://www.wolframalpha.com/input/?i=parabola+-16x%3Dy^2

anonymous · Answer

thank you but i noticed that this was not an option for my answer

anonymous · Answer

should be some form of that for sure

anonymous · Answer

like 
$$16x=y^2$$ or 
$$x=-\frac{1}{16}y^2$$ or something

anonymous · Answer

the options that i have are 

y^2 = -8x

 16y = x^2

 y = -(1/16)x^2

 x = -(1/16)y^2

anonymous · Answer

ah yes i just saw your other answer

anonymous · Answer

last one is the same as what we wrote

anonymous · Answer

i want to thank you so much for helping me

anonymous · Answer

yw

anonymous · Answer

i am going another one now if you want to see it
this one is a hyperbola

anonymous · Answer

ah yes that would be nice

anonymous · Answer

http://openstudy.com/study#/updates/533cbbcfe4b0fb0828593dda