math - elipse

[cool dude]

i have to teach myself conics, which isn't the easiest thing to do but i get most of it. i have a question though which i'm kinda stuck on, so i hope someone will know this.

write an equation in standard form for the ellipse.
foci at (0,0) and (0,8), and sum of focal radii 10

i did all the calculations and i'm pretty sure they're right because i checked it over and over because i kept thinking i might have screwed up somewhere. so in the final answer i got

25x^2/81 + y^2/9 = 1

this is however not exactly in standard form because the answer in the book says that the answer is

(x-4)^2/25 + y^2/9 = 1

i have the second part with the y correct but how do i reduce (i guess) the X

P.S. if u guys don't know how can u just tell me the method u would use to try and solve a question like this?

the method that i used was PF1 + PF2 = sum of focal radii

[Cervantes]

c = 4
a = 5
b = 3 (by b^2 = a^2 - c^2)

The general equation for an ellipse centred at the origin is:

x^2 / a^2 + y^2 / b^2 = 1

But in this question, the origin is not at the centre of the ellipse. The centre of the ellipse is (4, 0), so we have to "shift the equation" 4 to the right..

(x - 4)^2 / a^2 + y^2 / b^2 = 1

Now plug in values for a^2 and b^2

(x - 4)^2 / 25 + y^2 / 9 = 1

[cool dude]

that is easy when u know wat a and b are but in my question it only states the foci points and the foci radii. look at the above question. since we don't know the a or b we cannot use your method.
the equation that i made is

sqr(x^2 + y^2) + sqr(x^2 + (y + 8)^2) = 10

once u reduce everything and get rid of the square roots u will get the answer

25x^2/81 + y^2/9 = 1

this is partially right except for the x because it has to be

(x-4)^2/25 + y^2/9 = 1

[Cervantes]

Sure you know what a,b, and c are.

c is half the distance between the foci: 4

a is the semimajor axis. Since we know the sum of the focal radii is 10, we can easily determine where the (horizontal) edges of the ellipse are. Imagine when the line from one focal point to the locus back to the other focal point is horizontal: you're line goes from (0, 0) to the far right edge of the ellipse then back to (8, 0). The total distance of this trip is 10. We know most of that distance is 8, giving us 2 length units remaining to make the trip from the right focal point, to the edge, and back. Thus, the distance from the right focal point to the right edge is half of 2, which is 1. Therefore, the distance from the centre (4, 0) to the leftmost (-1, 0) or rightmost (9, 0) points on the ellipse is 5.

b is (a^2 - c^2)^0.5, which is 3.

cool dude wrote:

sqr(x^2 + y^2) + sqr(x^2 + (y + 8)^2) = 10

I don't see why you added 8 to y. Shouldn't it be more like

sqrt( x^2 + y^2 ) + sqrt( (x - 8)^2 + y^2) = 10

Though it has been a while since I've done these. Still, going from the definition of an ellipse, I get that.

[cool dude]

thanks so much u clarified a lot for me. the only thing i don't get is if the foci points are (0,0) and (0,8) isn't that supposed to be on the y axis?
and wat is c (i thought it was the point of the foci)?

***Edit*** also since its the y axis that has been slided up by four from the origin shouldn't it be (y - 4) instead of (x-4)

[Cervantes]

cool dude wrote:

and wat is c?

Cervantes wrote:

c is half the distance between the foci: 4

Raising to the power of by (1/2) is taking the sqaure root. Just like raising to the power of (1/3) is like taking the cube root

[cool dude]

that just confused me. then wat would i do if they told me wat a and b are and said find the foci points. wat i always did was do
a^2 - b^2 = c^2.

watever c was is the foci points?

that always worked but now u say that c is actually half the distance between the foci. so how would i get the foci points then?

***Edit*** why are we saying it moved 4 to the right when it moved 4 up from the origin. u are right though because thats wat the book says too but i don't understand why if the foci points are on the y axis and all of this was on the y axis

[Cervantes]

Once you know c, the foci c units away from the centre of the ellipse, along the major axis.

As for the second foci being (0, 8) instead of (8, 0): One of us is dyslexic.

[cool dude]

no the foci points are as written and no i'm not dyslexic. the foci points are (0,0) and (0,8). i get everyting now except for why is the right answer says in the book

(x-4)^2/25 + y^2/9 = 1

and i'm getting

x^2/25 + (y-4)^2/9 = 1

the way i see it is that the origin was moved on the y axis 4 up but according to the answer in the book if it is correct it clearly states that the origin is moved 4 to the right

***Edit*** P.S. in a hyperbola wat does it mean by difference of foci radi

[Cervantes]

So the answer in the back of the book is wrong. It's not like it's never happened before.

cool dude wrote:

***Edit*** P.S. in a hyperbola wat does it mean by difference of foci radi

The distance from one focal point to a point on the hyperbola minus the distance from the other focal point to the same point on the hyperbola is a constant.

If that's not good enough, try to Google for hyperbola. See how easy this is?

[cool dude]

so how can i get a, b and c with the following question

find equation of hyperbola centered at origin
foci at (-5,0) and (5,0) with the difference between the focal radii 6

[Cervantes]

Hey, how about you just put all your homework up and once and I'll do it in one go?

[cool dude]

sounds good although i have to learn myself i'm the one having a test. and no this is not homework, i am just trying to teach myself conics as stated clearly at the top of my post. also i asked u two main questions and u really don't have to answer them if u don't want to, but u also don't have to be rude.

P.S. even though your rudeness is not appreciated i do appreciate the help u gave me so far so thanks

[Cervantes]

My apologies for the brash tone.
I've given you a good resource, and I'm sure the textbook has lots of examples.

As for your question, the distance between the focal points is 10, and the difference in the focal radii is 6. From this, we can deduce that the edge of the hyperbola is 2 units in from the focal points. (Look at the horiz. line connecting the two focal points.)

c is the distance from the centre to either of the focal points: 5
a is the distance from the centre to either of the edges of the hyperbola. Put another way, a is half the width of the rectangle that you can draw. a is 3 (since the edges are 2 units from the focal points, as deduced above)
b is half the vertical distance of the rectangle. b^2 = c^2 - a^2, so b = 4

Now, the general equation is:


Plugging in our values, we get

x^2 / 9 - y^2 / 16 = 1

I think that will get you a hyperbola that opens sideways, though you'd want to check that. (If it doesn't open sideways, it opens up-down)

[cool dude]

thanks for the help i owe u one!!! i understand it now thanks to u

i don't think u care about bits but i'm still gonna donate some to u for your help. thanks again