Help Please! Will fan & medal!!
Write the general form for the equation of the following hyperbola: (x-1)^2/9 - (y-3)^2/4 = 1
x^2 - y^2 - 32x - 96y - 113 = 0 4x^2 - 9y^2 -4x + 108y - 55 = 0 4x^2 - 9y^2 -8x + 54y - 113 = 0 x^2 - y^2 - 8x + 54y - 113 = 0
Those are answers !
First step, multiply both sides by 9 and 4, and simplify it a bit.
After that, expand the squares. Then distribute the 4 and 9 through their groups. At this point, you'll be able to look at the x^2 term and the x term to get your answer. Only one choice has the same number in front of those two (the coefficient).
However, to do it completely, you'd just simplify, and then subtract 36 from both sides.
Can you show me please
Any particular part?
9 *(x-1)^2/9 * - (y-3)^2/4 = 9
(x-1)^2 * - (y-3)^2/4 = 9
(x-1)^2 * - 4(y-3)^2/4 = 9* 4 (x-1)^2 * - (y-3)^2 = 36 (x-1)^2 * - (y-3)^2 -36 = 1
like that^?
Close, very close! When you multiply by 9, the y term is also multiplied by 9. \[(x-1)^2 - 9*\frac{(y-3)^2}{4} = 9\]
Apply the same principle when you multiply by 4.
okay thats what i thought but it doesnt divide
That's ok. It'll work out well in just another step or so.
hmm so 4(x-1)^2 * - 36(y-3)^2 =36 ?
What happened to the 4 underneath the y?
4(x-1)^2 * - 9(y-3)^2 =36
canceled out ?
There you go :)
Now, expand the (x-1)^2 and the (y-3)^2
4(x-2x+1) * - 9(y-3y+ =36
idk how to complete the square for that second part
Very close! \[(x-1)^2 = (x-1)(x-1) = x^2 - 2x + 1\] \[(y-3)^2 = (y-3)(y-3) = y^2 - 6y + 9\] If I may ask, why were you completing the square?
Oops ugh its cuz im not use to solving these ones im use to solving something in general form to standard form haha
Now, distribute the 9 and 4 to their groups. Then you'll be able to see the answer :)
But okay so 4(x-2x+1) * - 9(y-6y+9) = 36 4x-8x2 - 9y+54y+81= 36 4x^2 - 9y^2 -8x + 54y - 113 = 0
??
yep! Good job!
Thanks!! :)) Hey i have another question do you mind me opening a new page and tagging you in it to help me or are you busy?
Small note, you messed up the sign on the 81 in the middle step, but it may have just been a typo. I can help with another question, sure.
yeah nice catch and okay! :)
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