Convert the equation to the standard form for a hyperbola by completing the square on x and y.

y^2- 25x^2 + 4y + 50x - 46 = 0

Question

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

y^2- 25x^2 + 4y + 50x - 46 = 0

zzr0ck3r · Answer

can you complete the square on $-25x^2+50x-46$?

zzr0ck3r · Answer

if not, can you find the vertex?

kimjafo · Answer

I tried before and got it wrong

zzr0ck3r · Answer

do you know how the vertex relates to completing the square?

kimjafo · Answer

No

zzr0ck3r · Answer

completing the square means to take a quadratic of the form $ax^2+bx+c$ to the form $a(x-h)^2+k$ where $(h,k)$ is the vertex.

Do you know what a vertex is?

kimjafo · Answer

I believe I do

zzr0ck3r · Answer

ok forget the vertex thing, we will just complete the square a different way. 

Using the vertex makes it easier but if you are not used to working with the vertex then I assume your teacher wants you to do it another way.

kimjafo · Answer

I do an online schooling, so I teach myself of time

zzr0ck3r · Answer

$y^2- 25x^2 + 4y + 50x - 46 = 0$

$y^2+4y- 25x^2 + 50x  = 46$

We need to complete the square twice for this problem. But one is harder than the other. 

$y^2+4v-25(x^2-2)=46$

$y^2+4y-25(x-1)^2=46-25(-1)^2$

$y^2+4y-25(x-1)^2=21$

What I did was factor out the -25 from the x terms and then complete the square for the quadratic in x

$(y+2)^2-25(x-1)^2=21+(-2)^2$

$(y+2)^2-25(x-1)^2=5^2$

zzr0ck3r · Answer

make sure that you understand that I completed the square twice and then I will show you how to do it.

kimjafo · Answer

Ah! I was doing the problems wrong 😅😅 thanks!