Question

Buenos dias alguien que me pueda ayudar porfavor

Answer 1

Respuesta:

$\frac{(x-3)^{2}}{16}+\frac{(y-1)^{2}}{4}=1$

Explicación paso a paso:

3x² + 12y² - 18x - 24y - 9 = 0

3x² - 18x + 12y² - 24y = 9

3(x² - 6x) + 12(y² - 2y) = 9

3(x² - 6x + 9) + 12(y² - 2y + 1) = 9 + 27 + 12

3(x - 3)² + 12(y - 1)² = 48

$\frac{3(x-3)^{2}}{48}+\frac{12(y-1)^{2}}{48}=\frac{48}{48}$

$\frac{(x-3)^{2}}{16}+\frac{(y-1)^{2}}{4}=1$

Answer 2

$3 {x}^{2} + 12 {y}^{2} - 18x - 24y - 9 = 0 \\ 3 {x}^{2} - 18x + 12 {y}^{2} - 24y = 9 \\3( {x}^{2} - 6x) + 12( {y}^{2} - 2y) = 9 \\ 3( {x}^{2} - 6x + 9) + 12( {y}^{2} - 2y + 1) = 9 \\ 3(x - 3)^{2} + 12 {(y - 1)}^{2} = 9 + 27 + 12 \\ 3 {(x - 3)}^{2} + 12 {(y - 1)}^{2} = 48 \\ \frac{3 {(x - 3)}^{2} }{48} + \frac{12 {(y - 1)}^{2} }{48} = \frac{48}{48} \\ \frac{ {(x - 3)}^{2} }{16} + \frac{ {(y - 1)}^{2} }{4} = 1$

Ahora el centro:

$C(3,1)$

Vertices:

$V1(3 + 4,1) \: V2(3 - 4,1)$

Focos:

${c}^{2} = {a}^{2} - {b}^{2} \\ c = \sqrt{16 - 4} \\ c = \sqrt{12} \\ c = 2 \sqrt{3}$

$F1(3 + 2 \sqrt{3} ,1) \: F2(3 - 2 \sqrt{3} ,1)$

Espero haberte ayudado :)

Éxitos.