find the focus and the directrix of the graph of each equation.
1.) y = 5x^2
2.) y = -1/8y^2

- pokiedokie

find the focus and the directrix of the graph of each equation.
1.) y = 5x^2
2.) y = -1/8y^2

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- pokiedokie

slight mistake on number 2: x = -1/8y^2

- Crystal_Bliss

you need to solve these?

- pokiedokie

yes but i'm confused on how and what equation i would use

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Crystal_Bliss

so um would you like me to slove both of them for you?.

- Crystal_Bliss

solve*

- pokiedokie

yes, but I'm feeling a bit confused so maybe you could go through the steps too?

- Crystal_Bliss

idk if i can help sorry it doesnt make sense

- pokiedokie

it's okay i understand

- Owlcoffee

These are parabolas whose foci is allocated inside the axis they represent, for instance, a parabola with the form: \(y= kx^2\) has it's foci allocated inside the y-axis, goes through the origin which means that the foci has to be same distant to the vertex (0,0) as the point that defines the intersection between the directrix and the y-axis.
A general form for the parabola whose foci is located in the y-axis, being a foci \(F(0,\frac{ p }{ 2 })\) and the directrix \(d)y=-\frac{ p }{ 2 }\) will yield as a result the following structure for a parabola:
\[y=\frac{ 1 }{ 2p }x^2\]
And look, compare it to \(y=5x^2\) the look exactly the same, and all we have to do is find that distance "p" in order to locate the points that matter, after all, the vertex is already in the origin so no need for complex calculations:
\[y=\frac{ 1 }{ 2p }x^2 \rightarrow y=5x^2 \iff \frac{ 1 }{ 2p }=5\]

- pokiedokie

so how do we find the distance? do i graph?

- Owlcoffee

Solve for "p"

- Owlcoffee

"p" is the distance from the foci to the directrix where the vertex is the mid-point.

- pokiedokie

ah okay, sorry

- Owlcoffee

What did you get as result?

- pokiedokie

i'm sorry i'm still a bit confused on how to go about this

- pokiedokie

maybe a similar example would help?

- Owlcoffee

Sure, let's for instance consider the equation of the parabola:
\[y=\frac{ 1 }{ 8 }x^2\]
And we desire to find the foci and the directrix of this parabola, well, we know the foci is located in the y-axis, because of the structure responding to \(y=\frac{ 1 }{ 2p }x^2\)
so what we will do is call \(\frac{ 1 }{ 2p }\) as equal to 1/8, therefore:
\[\frac{ 1 }{ 2p }=\frac{ 1 }{ 8 } \iff 2p=8 \iff p=4\]
"p" is the distance from the foci to the directrix, and we know the vertex is the mid point of this parabola, and the VERTEX is located in the origin, meaning (0,0).
So therefore, the foci must be located p/2 distance on the positive axis (because the coefficient of x^2 is positive) so therefore the foci must have coordinates: \(F(0,2)\) and the directrix, must be located in the negative part of the y-axis, and since the directrix is a line, in this case, parallel to the x-axis, we will know that it intersects the y-axis in the negative quadrant at -2, therefore: \(d)y=-2\).
So, now we know the foci and the directrix of the given equation of a parabola, those being \(F(0,2)\) and \(d)y=-2\) respectively.
|dw:1450146171422:dw|

- pokiedokie

ohh okay, thank you so much this explains a lot

- pokiedokie

so p = 2.5 for y=5x^2?

- pokiedokie

and if i'm not confused, the foci would be (0,1.25)?

- pokiedokie

sorry if i'm screwing it up

- Owlcoffee

Incorrect, from \[\frac{ 1 }{ 2p }=5\]
You won't get as a result p=2,5

- pokiedokie

oh okay

- pokiedokie

2p = 5 and it's not 2.5, then that means that it's just 5 and basically already solved?

- Owlcoffee

incorrect,
\[\frac{ 1 }{ 2p }=5 \]
Does not translate into \(2p=5\) since it's a proportion you corss multiply:
\[\frac{ 1 }{ 2p }=5 \iff (5)(2p)=1\]

- pokiedokie

ah okay, so 10p for cross multiplying

- Owlcoffee

good, solve for p there from.

- pokiedokie

F = (0,5) because of p/2, and the directrix would be y = -5?

- Owlcoffee

Incorrect, "5" is not the distance "p", when I showed you:
\[10p=1\]
this is not solved for "p", you can solve for "p" by diiding both sides by "10":
\[p=\frac{ 1 }{ 10 } \]
And that is the distance "p".

- pokiedokie

oh okay, very sorry i'm sorta new to this and i have difficulty with math

- Owlcoffee

okay, now that we know "p", we can do the whole, naming the foci and h directrix thing.

- pokiedokie

so how would we do that with a fraction? the fraction being 1/10

- Owlcoffee

Well, you are working with parabola so I have to assume you know how to operate fractions at this point, 1/10 divided 2 is the same as multiplying by 1/2.
\[\frac{ 1 }{ 10 }.\frac{ 1 }{ 2 }=\frac{ 1 }{ 20 }\]
And that is the distance to the foci and to the direcrix.
Regardless if it's a fracton or not, the process is the same.
which is first calculating "p" and then locating the foci and directrix.

- pokiedokie

okay, that makes a little more sense

- Owlcoffee

So, with all that you can locate the foci and the directrix.

Looking for something else?

Not the answer you are looking for? Search for more explanations.