anonymous
  • anonymous
I'm new to this website and how it works, but I would appreciate it if somebody helped me! Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

PhantomCrow
  • PhantomCrow
What did you notice when you graphed these two points on a plane?
anonymous
  • anonymous
@Phantomcrow It opens to the right?
PhantomCrow
  • PhantomCrow
Yes! And you know what the standard equation of a parabola that opens to the right is?

Looking for something else?

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

More answers

anonymous
  • anonymous
@Phantomcrow Can you help me out with that one
anonymous
  • anonymous
It is 3, you're welcome.
PhantomCrow
  • PhantomCrow
Sure.|dw:1444091089825:dw|
PhantomCrow
  • PhantomCrow
Since the vertex is at the origin, there is no need to add any constants for shifts. The graph is simply \[y^2=x\]
anonymous
  • anonymous
@Phantomcrow By the way the Answer choices are y= 1/28x^2 x= 1/28y^2 -28y= x^2 y^2=14x What confuses me is how these tie to the standard formula
PhantomCrow
  • PhantomCrow
D. Despite it having an 'x' term, its vertex is still at the origin.
PhantomCrow
  • PhantomCrow
Standard form is one of the ways to write the graph of a parabola. It is commonly seen as:\[ax^2+bx+c\]
anonymous
  • anonymous
@Phantomcrow Wow thanks, would it be cool if you helped me with one more?
PhantomCrow
  • PhantomCrow
Sure
anonymous
  • anonymous
@Phantomcrow Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8, i'm guessing this is similar or even exact to the last one, so it opens to the left
PhantomCrow
  • PhantomCrow
Actually, it does not open to the left this time. You are given y values for focus and directrix, not x values.
anonymous
  • anonymous
@Phantomcrow So y values open to the right and x values open to the left?
PhantomCrow
  • PhantomCrow
Y values lie across a vertical line, you your parabola will be opening upwards. You can easier see this if you look at the (0.8) and y= -8 on a graph. See that the they are equidistant from a point.
anonymous
  • anonymous
@PhantomCrow you graph (0,8) correct?
PhantomCrow
  • PhantomCrow
Yes. Graph the point (0,8) and y= -8 and you should be able to visually see the the point that is equidistant from them. There is a formula for finding distance but I believe simply looking at the graph for this problem will provide you with the answer.
anonymous
  • anonymous
@PhantomCrow The choices are y= 1/32x^2 y^2= 8x y^2=32x y=1/8x^2 Would it be B?
PhantomCrow
  • PhantomCrow
Oh, I see.
PhantomCrow
  • PhantomCrow
Well, you have two possible answers. A and D. B and C are ruled out because that would be a parabola who's focus is (x,0) and directrix is x=something. We have a focus like (0,y) and y=something in this case so its not y^2=x
PhantomCrow
  • PhantomCrow
Since you need to figure out which is correct between A and D, your best method is to plug in what you know into vertex form (\[(x-h)^2=4p(y-k)\] where (h, k) is the vertex and p is the distance from either the directrix or the focus to the vertex.
PhantomCrow
  • PhantomCrow
We know that the vertex lies at (0,0) because that is the point equidistant from (0,8) and y=-8. P is not difficult to find. P is just 8 because the distance from either the directrix or the focus to the vertex is 8 units (if you need the formula, just ask, but its not necessary).
PhantomCrow
  • PhantomCrow
Plug that into the vertex form:\[(x-h)^2=4p(y-k)\] \[(h,k)=(0,0)\] so \[h=0, k=0\] \[(x-0)^2=4(8)(y-k)\] \[x^2=32y\] \[\frac{ x^2 }{ 32 }=y\]
PhantomCrow
  • PhantomCrow
We want it in terms of y because that's how standard form is written. So we solved for y.
anonymous
  • anonymous
@PhantomCrow Which will result in y=1/32x^2
PhantomCrow
  • PhantomCrow
Yep.
anonymous
  • anonymous
@PhantomCrow Thank you so much for your help have a blessed day!
PhantomCrow
  • PhantomCrow
No problem, you as well.

Looking for something else?

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