Classify the conic and write the equation in standard form. 8x^2-25y+32x+150y-393=0
I know that its a hyperbola so we plug it in for that equation...

Question

Classify the conic and write the equation in standard form. 8x^2-25y+32x+150y-393=0
I know that its a hyperbola so we plug it in for that equation...

ammarah · Answer

@ganeshie8

ganeshie8 · Answer

that would be a mistake

ganeshie8 · Answer

its not an hyperbola

ganeshie8 · Answer

hyperbola will have both  $\large x^2$ and $\large  y^2$ terms, right ?

ganeshie8 · Answer

look at the equation,
u dont have any  $\large y^2$ terms

ammarah · Answer

im sorry its y^2 also

ammarah · Answer

8x^2-25y^2+32x+150y-393=0

ganeshie8 · Answer

Ohk... then you're right :) its a hyperbola !

ganeshie8 · Answer

u knw the standard form right ?

ammarah · Answer

yes

ganeshie8 · Answer

$\large   \dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$

ganeshie8 · Answer

its going to be very easy... 
u just need to complete the squre....

ganeshie8 · Answer

step1 : group x terms, group y terms,
and slap the constant to the other side

ganeshie8 · Answer

$\large    8x^2-25y^2+32x+150y-393=0$

group same variable terms :

$\large    8x^2+32x-25y^2 + 150y = 393$

ammarah · Answer

ok i got 8(x^2+4) - 25(y^2+150y)

ganeshie8 · Answer

looks good, except for mistake in sign : 

you should get :

$\large    8x^2-25y^2+32x+150y-393=0$

group same variable terms :

$\large    8x^2+32x-25y^2 + 150y = 393$


$\large    8(x^2+4x)-25(y^2 - 6y) = 393$

ammarah · Answer

yes

ganeshie8 · Answer

next complete the square for the stuff inside parenthesis by adding and subtracting the square term :


$\large    8x^2-25y^2+32x+150y-393=0$

group same variable terms :

$\large    8x^2+32x-25y^2 + 150y = 393$


$\large    8(x^2+4x)-25(y^2 - 6y) = 393$

$\large    8(x^2+4x + 2^2 - 2^2)-25(y^2 - 6y + 3^2 -3^2) = 393$

complete the square :

$\large    8((x+2)^2 - 2^2)-25((y-3)^2 -3^2) = 393$

ganeshie8 · Answer

see if that still looks fine :)

ammarah · Answer

yes an now we add the squares to the other side?

ammarah · Answer

oh wait noo its a hyperbola

ganeshie8 · Answer

exactly !


$\large    8x^2-25y^2+32x+150y-393=0$

group same variable terms :

$\large    8x^2+32x-25y^2 + 150y = 393$


$\large    8(x^2+4x)-25(y^2 - 6y) = 393$

$\large    8(x^2+4x + 2^2 - 2^2)-25(y^2 - 6y + 3^2 -3^2) = 393$

complete the square :

$\large    8((x+2)^2 - 2^2)-25((y-3)^2 -3^2) = 393$

$\large    8(x+2)^2 - 8*2^2-25(y-3)^2 +25*3^2 = 393$

$\large    8(x+2)^2 - 32-25(y-3)^2 + 225 = 393$


$\large    8(x+2)^2 -25(y-3)^2 + 193= 393$

$\large    8(x+2)^2 -25(y-3)^2 = 200$

ganeshie8 · Answer

see if you're fine so far

ganeshie8 · Answer

lastly, divide 200 both sides

ganeshie8 · Answer

$\large    8(x+2)^2 -25(y-3)^2 = 200$

divide 200 both sides :
$\large   \dfrac{ 8(x+2)^2}{200} -\dfrac{25(y-3)^2}{200} = 1$


simplify :
$\large   \dfrac{ (x+2)^2}{25} -\dfrac{(y-3)^2}{8} = 1$

ganeshie8 · Answer

we're done.

ammarah · Answer

ok yess im sorry to trouble u more but im stuck with writing equations of the graphs...

ganeshie8 · Answer

np :) i have time... go ahead