Need to get into standard form.

4y^2+x^2+7y+2x-6=-3

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- Jul 8th 2018, 06:25 PMNfalconerStandard form help
Need to get into standard form.

4y^2+x^2+7y+2x-6=-3 - Jul 8th 2018, 06:51 PMMarkFLRe: Standard form help
Try completing the square on x and y...what do you get?

- Jul 8th 2018, 07:18 PMNfalconerRe: Standard form help
All I got so far is (×+1)^2 + (4y^2 + 7y + 12 1/4)= 16 1/4. I am stuck there.

- Jul 8th 2018, 07:36 PMMarkFLRe: Standard form help
We are given:

$\displaystyle 4y^2+x^2+7y+2x-6=-3$

The first thing I would do is add 6 to both sides:

$\displaystyle 4y^2+x^2+7y+2x=3$

Next group by like variables on the left:

$\displaystyle \left(4y^2+7y\right)+\left(x^2+2x\right)=3$

We see $\displaystyle y^2$ has a coefficient other than 1, so factor that out:

$\displaystyle 4\left(y^2+\frac{7}{4}y\right)+\left(x^2+2x\right) =3$

Now add the square of half the coefficients of the linear terms (on both sides):

$\displaystyle 4\left(y^2+\frac{7}{4}y+\frac{49}{64}\right)+\left (x^2+2x+1\right)=3+4\cdot\frac{49}{64}+1$

Rewrite:

$\displaystyle 4\left(y+\frac{7}{8}\right)^2+(x+1)^2=\frac{113}{1 6}$

Multiply through by $\displaystyle \frac{16}{113}$

$\displaystyle \frac{64}{113}\left(y+\frac{7}{8}\right)^2+\frac{1 6}{113}(x+1)^2=1$

Write in standard form:

$\displaystyle \frac{(x+1)^2}{ \left(\dfrac{\sqrt{113}}{4}\right)^2}+ \frac{\left(y+\dfrac{7}{8}\right)^2}{ \left(\dfrac{\sqrt{113}}{8}\right)^2}=1$