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#### stephensnoah1

##### New member

- Jul 22, 2019

- 1

y^2+12y+16x+68=0

The form we have been using is (y-k)^2=4p(x-h)

Any explanation would help too, I'm really stuck on this one.

Thank you!

- Thread starter stephensnoah1
- Start date

- Thread starter
- #1

- Jul 22, 2019

- 1

y^2+12y+16x+68=0

The form we have been using is (y-k)^2=4p(x-h)

Any explanation would help too, I'm really stuck on this one.

Thank you!

- Apr 22, 2018

- 251

$$16x\ =\ 68-12y-y^2\ =\ 104-(y+6)^2$$

whence

$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$

in the form you have been using.