Did I do this coversion to Standard correctly?

1. Mar 16, 2005

trigger352

Cover to standard form...

$$4x^2 - 9y^2 + 32x + 18y + 91 = 0$$

$$4x^2 + 32x$$..................$$-9y^2 + 18y$$..................$$+91 = 0$$

$$4(x^2 + 8x + 16)$$..........$$-9(y^2 - 2y + 1)$$..........$$+91 -16 -1 = 0$$

$$4(x + 4)^2$$..................$$-9(y - 1)^2$$..................$$+74 = 0$$

therefore

$$\frac{(y - 1)^2}{4} - \frac{(x+4)^2}{9} = 74$$

2. Mar 16, 2005

Data

You made a few errors with the constant terms. See if you can figure out where.

3. Mar 16, 2005

trigger352

When the 74 goes over the = it becomes a negative, which is not possible. So, I'm thinking i did something wrong by dividing out -9. Somehow I need to get a balancing number that is larger that 91 so it won't be positive on the other side...
But I do see something that might work...

I have to change those balancing numbers. I only countered the -16 and -1. But those weren't multiplied by that factor.
Ok, So, $$91 - 64 + 9 = 36$$. 36?
But I can't get a positive number...since when it goes over the = line it'll be negative.

Last edited: Mar 16, 2005
4. Mar 16, 2005

Data

36 is correct. The negative is no problem: just multiply both sides by -1.

You made one more mistake though. To get the 4 and 9 in the denominators you had to divide the equation by 36. Do you see anything that you missed?