anonymous one year ago 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.

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1. PhantomCrow

What did you notice when you graphed these two points on a plane?

2. anonymous

@Phantomcrow It opens to the right?

3. PhantomCrow

Yes! And you know what the standard equation of a parabola that opens to the right is?

4. anonymous

@Phantomcrow Can you help me out with that one

5. anonymous

It is 3, you're welcome.

6. PhantomCrow

Sure.|dw:1444091089825:dw|

7. PhantomCrow

Since the vertex is at the origin, there is no need to add any constants for shifts. The graph is simply $y^2=x$

8. 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

9. PhantomCrow

D. Despite it having an 'x' term, its vertex is still at the origin.

10. PhantomCrow

Standard form is one of the ways to write the graph of a parabola. It is commonly seen as:$ax^2+bx+c$

11. anonymous

@Phantomcrow Wow thanks, would it be cool if you helped me with one more?

12. PhantomCrow

Sure

13. 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

14. PhantomCrow

Actually, it does not open to the left this time. You are given y values for focus and directrix, not x values.

15. anonymous

@Phantomcrow So y values open to the right and x values open to the left?

16. 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.

17. anonymous

@PhantomCrow you graph (0,8) correct?

18. 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.

19. anonymous

@PhantomCrow The choices are y= 1/32x^2 y^2= 8x y^2=32x y=1/8x^2 Would it be B?

20. PhantomCrow

Oh, I see.

21. 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

22. 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.

23. 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).

24. PhantomCrow

Plug that into the vertex form:$(x-h)^2=4p(y-k)$ $(h,k)=(0,0)$ so $h=0, k=0$ $(x-0)^2=4(8)(y-k)$ $x^2=32y$ $\frac{ x^2 }{ 32 }=y$

25. PhantomCrow

We want it in terms of y because that's how standard form is written. So we solved for y.

26. anonymous

@PhantomCrow Which will result in y=1/32x^2

27. PhantomCrow

Yep.

28. anonymous

@PhantomCrow Thank you so much for your help have a blessed day!

29. PhantomCrow

No problem, you as well.