# Find vertex, focus, and directrix of parabola: y^2+12y+16x+68=0

#### stephensnoah1

##### New member
Find the vertex, focus, and directrix of the following parabola:

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!

#### Olinguito

##### Well-known member
Hint: Rewrite the equation of the parabola as
$$16x\ =\ 68-12y-y^2\ =\ 104-(y+6)^2$$
whence
$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$
in the form you have been using.

#### Country Boy

##### Well-known member
MHB Math Helper
In other words "complete the square".

#### jonah

##### New member
Beer soaked correction follows.
$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$
When I saw this a month ago, I plotted it with my graphing app along with the OP's y^2+12y+16x+68=0. They didn't match. I figured somebody will come along with a correction but nobody did. Forgot about it till I decided to remove a few open tabs and saw it again. I guess nobody here is no longer double checking.

That should have been
$$(y+6)^2=4(-4)(x+2)$$

Edit: Just realized something else. This thread is a year old already. Country Boy's post made me think it was just new a month ago.

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#### Country Boy

##### Well-known member
MHB Math Helper
"Completing the square" as I suggested above:

A "perfect square" is of the form $(y+ a)^2= y^2+ 2ay+ a^2$. Comparing $y^2+ 2ay$ with $y^2+ 12y$ we must have $2ay= 12y$ so a= 6 and then $a^2= 36$. We need to add 36 to make this a "perfect square".

We can't just add a number to an expression and have the same numerical value but we can add and subtract the same number: $y^2+ 12y= y^2+ 12y+ 36-36= (y+ 6)^2- 36$.

So $y^2+12y+16x+68= (y+ 6)^2- 36+ 16x+ 68= (y+ 6)^2+ 16x+ 32= 0$.

$16x= -(y+6)^2- 32$

$x= -\frac{1}{16}(y+ 6)^2- 32$.

Now, that is a parabola with horizontal axis (parallel to the x-axis), opening to the left. Since a square in never negative, that is $-32$ minus something. x will be largest when $(y+ 6)^2= 0$, y= 6, where x= -32. The vertex is at (-32, -6).

When y= 0, $x= -\frac{1}{16}(0- 6)^2- 32= -\frac{9}{4}- 32= -34.25$. The x- intercept is at (-34.25, 0).

Since the leading coefficient, $-\frac{1}{16}$, is negative, the parabola opens to the left and never crosses the y- axis. There is no y- intercept.

#### jonah

##### New member
Beer soaked correction follows.
$x= -\frac{1}{16}(y+ 6)^2- 32$.
That should have been
$x= -\frac{1}{16}(y+ 6)^2- \frac{32}{16}$

#### Country Boy

##### Well-known member
MHB Math Helper
And $\frac{32}{16}= 2$! Yes, thank you.

#### jonah

##### New member
Tequila soaked ramblings follow.
And $\frac{32}{16}= 2$! Yes, thank you.
It need not be said but this of course invalidates some the rest after
$x= -\frac{1}{16}(y+ 6)^2- 32$

Cheers.

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