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

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Answer 1 · 2015-09-24T22:29:03

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

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

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

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

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

and

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

The equation may be rewritten

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

That is

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

Rearranging

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

$\Rightarrow 4^{2} {(x + 2)}^{2} + 3^{2} {(y + 6)}^{2} = 12^{2}$ $=> 4^2 (x + 2)^2 + 3^2(y + 6)^2 = 12^2$

$\Rightarrow \frac{4^{2}}{12^{2}} {(x + 2)}^{2} + \frac{3^{2}}{12^{2}} {(y + 6)}^{2} = 1$ $=> 4^2/12^2 (x + 2)^2 + 3^2/12^2(y + 6)^2 = 1$

$\Rightarrow {(\frac{4}{12})}^{2} {(x + 2)}^{2} + {(\frac{3}{12})}^{2} {(y + 6)}^{2} = 1$ $=> (4/12)^2 (x + 2)^2 + (3/12)^2(y + 6)^2 = 1$

$\Rightarrow {(\frac{1}{3})}^{2} {(x + 2)}^{2} + {(\frac{1}{4})}^{2} {(y + 6)}^{2} = 1$ $=> (1/3)^2 (x + 2)^2 + (1/4)^2(y + 6)^2 = 1$

$\Rightarrow \frac{{(x + 2)}^{2}}{3^{2}} + \frac{{(y + 6)}^{2}}{4^{2}} = 1$ $=> (x + 2)^2 / 3^2 + (y + 6)^2 / 4^2 = 1$

From which it may be noted:

the respective hemiradii are $3$ $3$ and $4$ $4$ (the square roots of the denominators of the terms in $x$ $x$ and $y$ $y$ , conventionally denoted by $b$ $b$ and $a$ $a$ , respectively, where $a$ $a$ conventionally denotes the major hemiradius (here, 4) and $b$ $b$ 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 $(- 2, - 6)$ $(-2, -6)$ (the negative of the constant terms added to $x$ $x$ and $y$ $y$ , respectively, to make "perfect" squares)

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

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

that is,

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

For this ellipse,

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

So, the foci are located at

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

and

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