Question

SAT subject test question:
the ellipse given by the equation 9x^2+4y^2-36x-12y+18=0 is centered at..
( I know that you need to do completing the square, but I don't know how to do that. so please show the steps! :) )

Answer 1

@zepdrix

Answer 2

@Callisto

Answer 3

\[\large 9x^2+4y^2-36x-12y+18=0\]So let's try to organize things a little, then we can take a shot at completing the squares.\[\large \color{green}{9x^2-36x}+\color{royalblue}{4y^2-12y}+18=0\]
Let's ignore the 18 for now.
Looking at the green terms, hmm let's factor out a 9 first.

Answer 4

\[\large \color{green}{9(x^2-4x)}+\color{royalblue}{4y^2-12y}+18=0\]

To complete the square, we take `half of the b term` and square it.
Assuming we're in this form \(\large x^2+bx\), which we are.

So taking half of our b term and squaring it gives us \(\large \left(\dfrac{-4}{2}\right)^2=4\)
This is the value that will complete the square on our x-terms.

\[\large \color{green}{9(x^2-4x+4)}+\color{royalblue}{4y^2-12y}+18=0\]But we can't just add 4 willy nilly, we have to keep the equation balanced, so we'll also subtract 4.\[\large \color{green}{9(\color{orangered}{x^2-4x+4}-4)}+\color{royalblue}{4y^2-12y}+18=0\]The part that I colored in orange is our perfect square.
So we need to get that -4 out of the brackets.
We do that by multiplying it by the 9.

\[\large \color{green}{9(\color{orangered}{x^2-4x+4})-36}+\color{royalblue}{4y^2-12y}+18=0\]

Answer 5

Is this explanation way too tedious? Do you already have some what of an understanding of completing the square? :o

Answer 6

I kind of.. I'm having a little trouble understanding this.
could you keep going until you get the final answer?

Answer 7

Being a perfect square, the x-terms will factor like so,
\[\large \color{green}{9(\color{orangered}{x-2})^2-36}+\color{royalblue}{4y^2-12y}+18=0\]If you do our math correctly, the term will always factor down to \(\large \left(x-\dfrac{b}{2}\right)^2\)

Answer 8

Completing the square can be difficult. It requires a bunch of little math tricks to get it just right.

We've completed the square on the x terms. Now let's try the y's.

Answer 9

Oh to be clear, you certainly don't have to do it this way.
But personally I find these a lot easier to deal with if you don't have a coefficient on the square term. That's why I like to factor that out if possible.

Answer 10

\[\large 9(x-2)^2+\color{royalblue}{4y^2-12y}+18-36=0\]So for these blues, let factor a 4 out of each y term.\[\large 9(x-2)^2+\color{royalblue}{4(y^2-3y)}-18=0\]

Answer 11

Again to complete the square, we'll take half of the b term and square it.

\(\large \left(\dfrac{-3}{2}\right)^2=\dfrac{9}{4}\)

Mmm ok factoring out the 4 doesn't work as well in this one...
It will still work out fine, but I have a feeling it will be more confusing to you since we're introducing a fraction. Hmm

Answer 12

Hmm oh well :( I'll just try to explain it this way, at least you can see the answer then. Harrtn probably has a better method for explaining this though.

\[\large 9(x-2)^2+\color{royalblue}{4(y^2-3y+\frac{9}{4}-\frac{9}{4})}-18=0\]So we'll add and subtract this term which completes the square for us.
Hopefully you understand this step.

Remember back in algebra when you could add something to both sides to keep the equation balanced? We're doing something a little different here, instead of adding to both sides, we're adding and subtracting the value at the same time. So we're essentially, in a sneaky way, adding zero.

Like we did with the x terms, we want to get rid of the -9/4. We'll multiply it by the 4 outside of the brackets.

\[\large 9(x-2)^2+\color{royalblue}{4(y^2-3y+\frac{9}{4})-9}-18=0\]
Now the part in the brackets will factor down to \(\large \left(y+\dfrac{b}{2}\right)^2\)

\[\large 9(x-2)^2+\color{royalblue}{4\left(y-\frac{3}{2}\right)^2-9}-18=0\]

Answer 13

Thank you so much!

Answer 14

Damn zepdrix, domination.

Answer 15

What do we need to do? Get it into ellipse form I guess? :o