Question

Word problemos -_-

Answer 1

For second question:

A second degree equation ax^2+2hxy+by^2+2gx+2fy+c=0

Becomes a parabola when h^2=ab
hyberbola when h^2>ab
ellipse when h^2<ab

Answer 2

@AravindG Thank you so much,your the best ;D

Answer 3

:)

Answer 4

@mary.rojas can you answer now lol!

Answer 5

Let me try the first problem

Answer 6

No xD

Answer 7

:O

Answer 8

But am sure it is useful information

Answer 9

lmao am sorry but that is a different language to me

Answer 10

hmm put the values as Aravind said! we will help do it here

Answer 11

okay,hold on,give me a sec!

Answer 12

k :)

Answer 13

Indeed first take all terms to one side. You get a second degree equation. Compare for h,a and b. Then evaluate h^2-ab

Answer 14

umm,so yeah,I dont know what am doing

Answer 15

Are you allowed to use a calculator?

Answer 16

yes

Answer 17

You dont need one lol ;)

Answer 18

Any observation?

Answer 19

Oh ;)
and what do you mean?......I can group the variables and factor?

Answer 20

can you identify a,b and h in the equation

Answer 21

ohhhhhhhhh yes, wait

Answer 22

No I mean whatever horrible number you see in the equation..Use your intellect to get an intelligent answer.

Answer 23

a=3
b=-4.84
h=I dont know

AravindG..... -_-

Answer 24

h is the xy term which is missing here so we can take it as zero

Answer 25

lol Identifying h was intelligent part lol XD

Answer 26

There is no xy term!! Means h=0 >.>

Answer 27

Where do you see an xy term? there is only a^2, x, y^2, and y????

Answer 28

@AravindG ;p

Answer 29

Oh wow, lol ok so h=0

Answer 30

Yeah there is no xy term which means h=0
0*(xy) =0

Answer 31

mhmmm now what?

Answer 32

Now can you identify which conic it represent? Its easy as a pie ;)

Answer 33

Now find value of h^2-4ab

Answer 34

I have h=0 --->h^2=0

Answer 35

@thisSucks h^2-ab

Answer 36

@AravindG are you sure?

Answer 37

have a look http://www.ck12.org/book/CK-12-Algebra-II-with-Trigonometry-Concepts/r4/section/10.11/

Answer 38

wellll.....if ts is right it is 58.08? and if AravindG is correct(which I think he ate to much pie to know ;) ) then it is 14.52?

Answer 39

*too

Answer 40

You see there is no calculation required here also. Just think like this:

I need a,b,h ..OK I found h=0
a=coeff of x^2 term, b=coeff of y^2 term.

@thisSucks I am sure.

Answer 41

So no pie for me :( anyway in both the case value of discriminant is +ve so we know it is hyperbola

Answer 42

*gives ts pie ;D
And hmm..okay

Answer 43

@mary.rojas You missed the second intelligent part :/

Answer 44

ARGGG WHAT?

Answer 45

@AravindG did you check the link.....don't let me go to the proof. I don't remember the conic very well but I trust Ck12

Answer 46

Yep there was no need to evaluate ab

a is positive sign b is negative. Product is negative

Means 0-(-STHNG)=+ve

So h^2>ab and hence hyperbola.

Answer 47

@mary.rojas getting confused lol

Answer 48

@_@ @thisSucks

Answer 49

@thisSucks I think I have this in control. Can you please move to another question?This is getting laggy :/

Answer 50

So Mary did u get it?

Answer 51

K @AravindG I will leave you guys but do check you fact. discriminant is b^2-4ac

Answer 52

*_*

Answer 53

@thisSucks What you learned is right. So is my knowledge. The only mistake you are making is applying it in the wrong place. Look carefully, in your equation and mine. Your equation has a Bxy term and not a 2hxy term. I hope you get it now :)

Answer 54

@AravindG Yes my equation just is of the form a^2 + b^2 +hxy...yeah so if you take care of 2 we are good!

Answer 55

:)

Answer 56

xD

Answer 57

WHO IS RIGHT??!?!?!

Answer 58

you do the second part...it depends on value of h you find!

Answer 59

OH GOODNESS GRACIOUS OKAY,HOLD ON

Answer 60

you follow the equation by arvind I leave now!

Answer 61

Post a new thread pls :/

Answer 62

AGG OKAY