pokiedokie
  • pokiedokie
find the focus and the directrix of the graph of each equation. 1.) y = 5x^2 2.) y = -1/8y^2
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
pokiedokie
  • pokiedokie
slight mistake on number 2: x = -1/8y^2
Crystal_Bliss
  • Crystal_Bliss
you need to solve these?
pokiedokie
  • pokiedokie
yes but i'm confused on how and what equation i would use

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Crystal_Bliss
  • Crystal_Bliss
so um would you like me to slove both of them for you?.
Crystal_Bliss
  • Crystal_Bliss
solve*
pokiedokie
  • pokiedokie
yes, but I'm feeling a bit confused so maybe you could go through the steps too?
Crystal_Bliss
  • Crystal_Bliss
idk if i can help sorry it doesnt make sense
pokiedokie
  • pokiedokie
it's okay i understand
Owlcoffee
  • 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
  • pokiedokie
so how do we find the distance? do i graph?
Owlcoffee
  • Owlcoffee
Solve for "p"
Owlcoffee
  • Owlcoffee
"p" is the distance from the foci to the directrix where the vertex is the mid-point.
pokiedokie
  • pokiedokie
ah okay, sorry
Owlcoffee
  • Owlcoffee
What did you get as result?
pokiedokie
  • pokiedokie
i'm sorry i'm still a bit confused on how to go about this
pokiedokie
  • pokiedokie
maybe a similar example would help?
Owlcoffee
  • 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
  • pokiedokie
ohh okay, thank you so much this explains a lot
pokiedokie
  • pokiedokie
so p = 2.5 for y=5x^2?
pokiedokie
  • pokiedokie
and if i'm not confused, the foci would be (0,1.25)?
pokiedokie
  • pokiedokie
sorry if i'm screwing it up
Owlcoffee
  • Owlcoffee
Incorrect, from \[\frac{ 1 }{ 2p }=5\] You won't get as a result p=2,5
pokiedokie
  • pokiedokie
oh okay
pokiedokie
  • pokiedokie
2p = 5 and it's not 2.5, then that means that it's just 5 and basically already solved?
Owlcoffee
  • 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
  • pokiedokie
ah okay, so 10p for cross multiplying
Owlcoffee
  • Owlcoffee
good, solve for p there from.
pokiedokie
  • pokiedokie
F = (0,5) because of p/2, and the directrix would be y = -5?
Owlcoffee
  • 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
  • pokiedokie
oh okay, very sorry i'm sorta new to this and i have difficulty with math
Owlcoffee
  • Owlcoffee
okay, now that we know "p", we can do the whole, naming the foci and h directrix thing.
pokiedokie
  • pokiedokie
so how would we do that with a fraction? the fraction being 1/10
Owlcoffee
  • 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
  • pokiedokie
okay, that makes a little more sense
Owlcoffee
  • Owlcoffee
So, with all that you can locate the foci and the directrix.

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