# Question #30429

May 9, 2017

The standard form is ${\left(x + 1\right)}^{2} / {\left(\sqrt{5}\right)}^{2} + {\left(y - 1\right)}^{2} / {4}^{2} = 1$

#### Explanation:

Let's rearrange the equation by completing the squares

$16 {x}^{2} + 5 {y}^{2} + 32 x - 10 y - 59 = 0$

$16 \left({x}^{2} + 2 x\right) + 5 \left({y}^{2} - 2 y\right) = 59$

$16 \left({x}^{2} + 2 x + 1\right) + 5 \left({y}^{2} - 2 y + 1\right) = 59 + 16 + 5$

$16 {\left(x + 1\right)}^{2} + 5 {\left(y - 1\right)}^{2} = 80$

$16 {\left(x + 1\right)}^{2} / 80 + 5 {\left(y - 1\right)}^{2} / 80 = 1$

${\left(x + 1\right)}^{2} / 5 + {\left(y - 1\right)}^{2} / 16 = 1$

${\left(x + 1\right)}^{2} / {\left(\sqrt{5}\right)}^{2} + {\left(y - 1\right)}^{2} / {4}^{2} = 1$

This is an ellipse, center $= \left(- 1 , 1\right)$

graph{(16x^2+32x+5y^2-10y-59)((x+1)^2+(y-1)^2-0.01)=0 [-11.25, 11.25, -5.625, 5.625]}

May 9, 2017

Yes

#### Explanation:

To convert the equation to standard form for ellipses, we need to complete the square.
Note that the standard form for an ellipse is:
$\frac{{\left(x - h\right)}^{2}}{a} ^ 2 + \frac{{\left(y - k\right)}^{2}}{b} ^ 2 = 1$ with a horizontal semi-major axis
or
$\frac{{\left(x - h\right)}^{2}}{b} ^ 2 + \frac{{\left(y - k\right)}^{2}}{a} ^ 2 = 1$ with a vertical semi-major axis
where $\left(h , k\right)$ is the center of the ellipse and $a > b$, where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis

Given the starting equation $16 {x}^{2} + 5 {y}^{2} + 32 x - 10 y - 59 = 0$, we first add 59 to both sides to isolate the variables and the constants:
$16 {x}^{2} + 5 {y}^{2} + 32 x - 10 y = 59$

Now we can complete the square for x and y:
$16 \left({x}^{2} + 2 x\right) + 5 \left({y}^{2} - 2 y\right) = 59$
$16 {\left(x + 1\right)}^{2} - 16 + 5 {\left(y - 1\right)}^{2} - 5 = 59$
$16 {\left(x + 1\right)}^{2} + 5 {\left(y - 1\right)}^{2} = 80$

Now we divide both sides by 80 to make the right-hand side equal to 1 for standard conic form:
$\frac{{\left(x + 1\right)}^{2}}{5} + \frac{{\left(y - 1\right)}^{2}}{16} = 1$

Therefore, this conic is a vertical ellipse centered at $\left(- 1 , 1\right)$ with a semi-major axis of length $4$ and a semi-minor axis of length $\sqrt{5}$.