# Homework Help: Finding points of intersection

1. Jun 18, 2007

### physicsgal

the parabola equation is:
(x^2/25) - (y^2/9) = 1

the line is y = 4-x

according to my calculations, if i point y - 4-x into the equation, i get
-16x^2 + 200x - 175 = 0. is that right so far?

~Amy

2. Jun 18, 2007

### VietDao29

No, it's not correct. =.="

Can you show us how you get there?

I get: -16x2 + 200x - 625 = 0.

Btw, that's definitely not a parabola equation. Instead, it's a hyperbola equation.

Last edited: Jun 18, 2007
3. Jun 18, 2007

### physicsgal

thanks

i meant to say hyperbola :shy:

i have 8 lines of work and my typing numbers is slow.. so here's part of my calculations:

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

225(x^2/25) - 225 ((4-x)^2/9) = 225(1)
9x^2 - 25(16 - 4x - 4x + x^2) = 225
9x^2 - 25(16 - 8x + x^2) = 225
9x^2 - 400 + 200x - 25x^2 = 225

i see my mistake now thanks

and then after i do the quadratic equation, i end up with x = 6.25, so y= -2.25. and that's my only intersection.

~Amy

4. Jun 18, 2007

### VietDao29

Yeah, looks good. Congratulations. :)

5. Jun 18, 2007

### physicsgal

if you dont mind.. one more hyperbola question:

4x^2 - y^2 + 8x + 4y + 16 = 0
what is this in standard form?

in standard form:
(x+1)^2/4 - (y-2)^2/16 = 1

does that look accurate?

~Amy

6. Jun 18, 2007

### malawi_glenn

You can try to expand that and see if you get the same as the initial form :)

7. Jun 18, 2007

### physicsgal

thanks. i expanded it and got:
4x^2 - y^2 + 8x + 4y + 4 = 0

so.. either something went wrong when i was expanding, or the standard form i figured out is incorrect?

~Amy

8. Jun 18, 2007

### malawi_glenn

you are very close altough, if you do it again properly and careful :) i promise.

9. Jun 18, 2007

### physicsgal

so there was something wrong with my expanding?

~Amy

10. Jun 18, 2007

### malawi_glenn

"if you dont mind.. one more hyperbola question:

4x^2 - y^2 + 8x + 4y + 16 = 0
what is this in standard form?"

You said, then you gave:

" (x+1)^2/4 - (y-2)^2/16 = 1"

as the normal form of the first equation you gave;
and when this is expand, you get:

4(x^2 + 2x + 1) - (y^2-4y + 4) = 16 ->
4x^2 + 8x + 4 - y^2 + 4y - 4 = 16 ->
4x^2 + 8x - y^2 + 4y - 16 = 0

right?

try:

-[(x+1)^2]/4 + [(y-2)^2]/16 = 1

expand:

-4[x^2 +2x+1] + [y^2 - 4y + 4] = 16 ->
4[x^2 +2x+1] - [y^2 - 4y + 4] +16 = 0 ->
4x^2 + 8x + 4 - y^2 + 4y - 4 + 16 = 0 ->
4x^2 - y^2 + 8x + 4y + 16 = 0

Last edited: Jun 18, 2007
11. Jun 18, 2007

### physicsgal

k thanks :)

but could i write that as:
[(x+1)^2]/4 + [(y-2)^2]/16 = -1

(i dont think its standard form to have a "-" infront of the equation)

and for this one, b^2 = 16 ?
(i have to do other calculations)
~Amy

12. Jun 19, 2007

### malawi_glenn

Well as you said, having a minus in front of the "1" is not standard form. So why do you want to do that?

13. Jun 19, 2007

### VietDao29

Nope, you can't. However, you "can" write it as:

[(x+1)^2]/4 - [(y-2)^2]/16 = -1

Yes, you can write it like that, but it's definitely not a standard form.

[(y-2)^2]/16 - [(x+1)^2]/4 = 1

Note that it's a North-south opening hyperbola, your b2 = 4. :)

Last edited: Jun 19, 2007
14. Jun 20, 2007

### physicsgal

thanks for the help guys ). im finished the course (just have to do the exam). :)

~Amy

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