Question

Resuelva y determine los elementos que conforma la circunferencia,

parábola y elipse con las siguientes ecuaciones:

2x2

+ 8x – y + 8 = 0

y

2

– 6y -8x + 17 = 0

x

2 + y

2

– 4x + 6y = 0

2x² + 3y² -8x +9y -6 = 0
aiudaaaaa​

Answer 1

Respuesta:

------------------------------------------

[tex]8x-y+8=0\\8x-y+8+y=0+y\\8x+8=y\\8x+8-8=y-8\\8x=y-8\\\frac{8x}{8}=\frac{y}{8}-\frac{8}{8}\\x=\frac{y-8}{8}[/tex]

------------------------------------------

[tex]-6y-8x+17=0\\-6y-8x+17+6y=0+6y\\-8x+17=6y\\-8x+17-17=6y-17\\-8x=6y-17\\\frac{-8x}{-8}=\frac{6y}{-8}-\frac{17}{-8}\\x=-\frac{6y-17}{8}[/tex]

------------------------------------------

[tex]-4x+6y=0\\-4x+6y-6y=0-6y\\-4x=-6y\\\frac{-4x}{-4}=\frac{-6y}{-4}\\x=\frac{3y}{2}[/tex]

------------------------------------------

[tex]2x^2+3y^2-8x+9y-6=0\\\\\frac{\left(x-2\right)^2}{\left(\frac{\sqrt{83}}{2\sqrt{2}}\right)^2}+\frac{\left(y-\left(-\frac{3}{2}\right)\right)^2}{\left(\frac{\sqrt{83}}{2\sqrt{3}}\right)^2}=1\\\\\left(h,\:k\right)=\left(2,\:-\frac{3}{2}\right),\:a=\frac{\sqrt{83}}{2\sqrt{2}},\:b=\frac{\sqrt{83}}{2\sqrt{3}}\\\\\left(h,\:k\right)=\left(2,\:-\frac{3}{2}\right),\:\:a=\frac{\sqrt{83}}{2\sqrt{2}},\:\:b=\frac{\sqrt{83}}{2\sqrt{3}}[/tex]

F<3