anonymous
  • anonymous
What are the foci of the hyperbola given by the equation 16x2 – 9y2 + 64x + 72y – 224 = 0 ? (–7, –4) and (3, –4) (–7, 4) and (3, 4) (–3, –4) and (7, –4) (–3, 4) and (7, 4)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
$$ 16x^2-9y^2+64x+72y-224=0\\ 16(x^2-2(2)(x)+ \color{red}{?^2})-9(y^2-2(4)(y)+\color{red}{?^2})-224=0 $$
jdoe0001
  • jdoe0001
what numbers there would complete the perfect square trinomial? @pvs285
anonymous
  • anonymous
help plz

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anonymous
  • anonymous
i need the answer only plz
rajee_sam
  • rajee_sam
if you complete the squares and write the equation of the hyperbola in standard form you get \[\frac{ (x+2)^{2} }{ 9 } - \frac{ ( y - 4)^{2} }{ 16 } = 1\]
jdoe0001
  • jdoe0001
$$ \frac{ (x+2)^{2} }{ 3^2 } - \frac{ ( y - 4)^{2} }{ 4^2 } = 1 $$ distance from the center (h,k) to the foci is $$ \sqrt{a^2+b^2} $$
anonymous
  • anonymous
ok...
jdoe0001
  • jdoe0001
bear in mind that the fraction with the "x" in it has the POSITIVE sign, not the negative, thus the hyperbola opens horizontally, so you move to the foci from the "h" value of the (h,k) center
jdoe0001
  • jdoe0001
$$ \text{so one focus is at}\\ h+\color{blue}{\sqrt{a^2+b^2}}\\ \text{and the other focus at}\\ h-\color{blue}{\sqrt{a^2+b^2}}\\ $$
anonymous
  • anonymous
whats the answer need it quaikly
jdoe0001
  • jdoe0001
dunno what's \(\sqrt{a^2+b^2} ?\)
rajee_sam
  • rajee_sam
now the center of the hyperbola is at (-2,4) and the hyperbola opens right and left. Since a = 3 and b= 4 the vertex will be 3 points away from the center. So the vertices are ( -5,4) and ( 1,4) Now to find the focus we need to do c² = a² + b² = 9 + 16 = 25. C² = 25; C = 5 So the focus is 5 points away from the center. Foci are ( -7,4) and ( 3,4) |dw:1370039439321:dw|
anonymous
  • anonymous
ty

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