Question

Use completing the square to graph 4x^2+8x+3y^2-18y=-15

Answer 1

Divide by 3 for both sides.

Answer 2

and then complete the square for y only first.

Answer 3

\[\frac{ 4 }{ 3 }x^2+\frac{ 8 }{ 3 }x+y^2-6y=-5\]

Answer 4

\[y^2-6y=-5\]
Ignore the x's for a moment and complete the square for this first.

Answer 5

y=5 and y=1

Answer 6

No...complete the square without finding the y values.

Answer 7

just find it in this form:
\[(y-a)^2=b+a^2\]

Answer 8

\[(y-3)^{2}=4\]

Answer 9

Put it in this form sorry. not find.

Answer 10

okay you just did the first part.
Now.
\[\frac{ 4 }{ 3 }x^2+\frac{ 8 }{ 3 }x+(y-3)^2=4\]

Answer 11

Now multiply by 3 to both sides.

Answer 12

\[4x^2+8x+3(y-3)^2=12\]

Answer 13

Now to complete the square the equation of the x's must be monic meaning it has to have no coefficient.

Answer 14

That means we have to divide by 4 now to get rid of the coefficient of x^2

Answer 15

You will get this result when you divide both sides by 4.
\[x^2+2x+\frac{3}{4}(y-3)^{2}=3\]

Answer 16

No don't worry about the
\["+\frac{3}{4}(y-3)^2"\]

Answer 17

Complete the square for x now as usual.
\[x^2+2x=3\]

Answer 18

@imnotatowel

Answer 19

\[(x+1)^{2}=4\]

Answer 20

Good. Now you combine the result and you get:
\[(x+1)^2+\frac{3}{4}(y-3)^2=4\]

Answer 21

multiply by 4 to both sides just to get rid of that fraction in the equation.

Answer 22

wait don't multiply just yet.

Answer 23

move the \[\frac{3}{4}(y-3)^2\] to the RHS.

Answer 24

RHS?

Answer 25

right hand side?

Answer 26

Wait. Nah, just multiply by 4 to both sides.

Answer 27

Forget about moving it the Right hand side,

Answer 28

moving it to*

Answer 29

Now you know how to graph that? If you don't know how to graph that, then just plot points and see where that leads you.

Answer 30

I know x+1 will shift it to the left one, but i don't know what the four does to that

Answer 31

It's a circle.

Answer 32

\[x^2+y^2=r^2\] where r is the radius.

Answer 33

with a radius of 4

Answer 34

\[(x-h)^2+(y-k)^2=r^2\] where (h, k) is the vertex.And r^2=4. r is the radius. so r is the square root of 4.

Answer 35

centre not vertex*

Answer 36

the centre*

Answer 37

How do i plot the centre?

Answer 38

Wait I figured it out, thanks

Answer 39

I'm not sure about the 4 and 3 beside the brackets. I might need to ask someone about that part. @AravindG Do you know how to draw this circle?
\[4(x+1)^2+3(y-3)^2=12\]
And sorry, the radius is sqrt(12) not 2. You forgot to multiply by 4 to boths sides.

Answer 40

both sides*

Answer 41

You sure you figured it out @imnotatowel ?

Answer 42

for the equation you wrote above we can divide the whole equation by 12 to get the form

\[(x-a)^2+(y-b)^2=r^2\]

Answer 43

no wouldn't it be fraction+fraction=1?

Answer 44

because if you divide by 12 to both sides, you get 1/3(x+1)^2 + 1/4(y-3)^2=1

Answer 45

oh yeah....gimme a minute

Answer 46

well actually the figure will be an ellipse!!

Answer 47

take a look:

http://www.wolframalpha.com/input/?i=4%28x%2B1%29%5E2%2B3%28y-3%29%5E2%3D12

Answer 48

Ah okay, I had an inkling about the numbers beside the brackets. I'm not up to ellipses yet but then again, this isn't my question so, thanks for helping @imnotatowel

Answer 49

infact general form of an ellipse is

\[\large \dfrac{(x-h)^2}{a^2}+\dfrac{(y-b)^2}{b^2}=1\]

where (h,k) is centre of the ellipse.Next time ensure you dont get that inky feeling again @Azteck :)

Answer 50

*inkling

Answer 51

lolz. Ahaha you can count on it.