# What are the foci of the ellipse given by the equation 16x^2 + 9y^2 + 64x + 108y + 244 = 0??

Sep 24, 2015

The foci are located at $\left(- 2 , \left(- 6 - \sqrt{7}\right)\right)$ and $\left(- 2 , \left(- 6 + \sqrt{7}\right)\right)$

#### Explanation:

$16 {x}^{2} + 9 {y}^{2} + 64 x + 108 y + 244 = 0$

may be rearranged (for convenience, to group terms variously in $x$ and $y$ to see what is happening) to:

$16 {x}^{2} + 64 x + 9 {y}^{2} + 108 y + 244 = 0$

Noting ("completing the square")

$16 {x}^{2} + 64 x = {\left(4 x + 8\right)}^{2} - 64 = {4}^{2} {\left(x + 2\right)}^{2} - 64$

and

$9 {y}^{2} + 108 y = {\left(3 y + 18\right)}^{2} - 324 = {3}^{2} {\left(y + 6\right)}^{2} - 324$

The equation may be rewritten

${4}^{2} {\left(x + 2\right)}^{2} - 64 + {3}^{2} {\left(y + 6\right)}^{2} - 324 + 244 = 0$

That is

${4}^{2} {\left(x + 2\right)}^{2} + {3}^{2} {\left(y + 6\right)}^{2} - 144 = 0$

Rearranging

${4}^{2} {\left(x + 2\right)}^{2} + {3}^{2} {\left(y + 6\right)}^{2} = 144$

$\implies {4}^{2} {\left(x + 2\right)}^{2} + {3}^{2} {\left(y + 6\right)}^{2} = {12}^{2}$

$\implies {4}^{2} / {12}^{2} {\left(x + 2\right)}^{2} + {3}^{2} / {12}^{2} {\left(y + 6\right)}^{2} = 1$

$\implies {\left(\frac{4}{12}\right)}^{2} {\left(x + 2\right)}^{2} + {\left(\frac{3}{12}\right)}^{2} {\left(y + 6\right)}^{2} = 1$

$\implies {\left(\frac{1}{3}\right)}^{2} {\left(x + 2\right)}^{2} + {\left(\frac{1}{4}\right)}^{2} {\left(y + 6\right)}^{2} = 1$

$\implies {\left(x + 2\right)}^{2} / {3}^{2} + {\left(y + 6\right)}^{2} / {4}^{2} = 1$

From which it may be noted:

• the respective hemiradii are $3$ and $4$ (the square roots of the denominators of the terms in $x$ and $y$, conventionally denoted by $b$ and $a$, respectively, where $a$ conventionally denotes the major hemiradius (here, 4) and $b$ typically denotes the minor hemiradius (here, 3)).
• the ellipse is not in canonical position as the major axis is aligned with the y-axis rather than with the x-axis and the ellipse is not centred on the origin.
• the centre of the ellipse is located at $\left(- 2 , - 6\right)$ (the negative of the constant terms added to $x$ and $y$, respectively, to make "perfect" squares)

Denoting the absolute value of the distance of the foci from the centre of the ellipse by $c$, it is noted

${c}^{2} = {a}^{2} - {b}^{2}$,

that is,

$c = \sqrt{{a}^{2} - {b}^{2}}$

For this ellipse,

$c = \sqrt{{4}^{2} - {3}^{2}} = \sqrt{16 - 9} = \sqrt{7}$

So, the foci are located at

$\left(- 2 , - 6 - \sqrt{7}\right)$

and

$\left(- 2 , - 6 + \sqrt{7}\right)$