## ksaimouli Group Title find intersection in terms of y of y^2-4x=4and 4x-y=16 one year ago one year ago

1. ksaimouli Group Title

$y^2-4x=4 and 4x-y=16$

2. ksaimouli Group Title

@ZeHanz

3. Hero Group Title

Hint: y = 4x - 16

4. ksaimouli Group Title

i did but i dont know whre i was wrong

5. ksaimouli Group Title

i know dude to solve for x and set it eual and then solve for x

6. ksaimouli Group Title

|dw:1357491125167:dw|

7. ksaimouli Group Title

|dw:1357491191134:dw|

8. Hero Group Title

Don't make it complicated bro

9. ksaimouli Group Title

|dw:1357491240591:dw|

10. ksaimouli Group Title

solve for y am i right this is to find area between curve sooo i must to with respect y

11. Hero Group Title

Actually, that's a good approach after all.

12. ksaimouli Group Title

but something is wrong in solving hmm i am not getting the answer

13. ksaimouli Group Title

multiply both sides by 4

14. Hero Group Title

What did you get for the answer?

15. ksaimouli Group Title

.08 and -12.08

16. Hero Group Title

That's what you got for y values?

17. ksaimouli Group Title

nope

18. ksaimouli Group Title

i gote it 5 and -4

19. Hero Group Title

Okay, so what's the problem?

20. ksaimouli Group Title

arithmetic mistake lol

21. ksaimouli Group Title

now i have to integrate thats it

22. ksaimouli Group Title

|dw:1357491558149:dw|

23. ksaimouli Group Title

but not sure which one goes top and which goes bottom with respect to y

24. Hero Group Title

Here's another way to do it to avoid confusion: $\frac{y^2 - 4}{4} - \frac{16 + y}{4} = 0$

25. ksaimouli Group Title

ya i tried that worked well

26. Hero Group Title

Bro, a is always the lowest value. b is always the highest value.

27. ksaimouli Group Title

i mean the function not the points (curves)

28. ksaimouli Group Title

$x=\frac{ y^2-4 }{ 4 }$

29. Hero Group Title

Are you sure you're supposed to be using y values for a and b?

30. ksaimouli Group Title

or$x=\frac{ 16+y }{ 4 }$

31. ksaimouli Group Title

ya i am damn sure

32. Hero Group Title

I remember these. They can be tricky. What you do is just turn your graph sideways. And it will be the same as like when you're working with x-axis.

33. ksaimouli Group Title

if $\sqrt{y}$

34. ksaimouli Group Title

|dw:1357491948129:dw|

35. ksaimouli Group Title

?

36. ksaimouli Group Title

alright thx

37. Hero Group Title

Bro, I graphed it as $\frac{16 + x}{4}$ and $\frac{x^2 - 4}{4}$ You get the same area when integrating from a = -4 to b = -5 The only difference is you're using variable y instead of variable x

38. Hero Group Title

@ksaimouli

39. ksaimouli Group Title

ya i noticed it thx