anonymous
  • anonymous
Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AlexandervonHumboldt2
  • AlexandervonHumboldt2
Focus: (2, 6) Any point, (x0 , y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate: Distance between the point on the parabola to the focus Distance between the point on the parabola to the directrix \[\sqrt{(x_0-a)^2+(y_0-b)^2} \] \[\sqrt{(x_0-2)^2+(y_0-0)^2}\] y=-2 dirextix Distance between point ( x0 , y0) and the line y=-2 |y_0+2| Equate the two expressions. \[\sqrt{(x_0-2)^2+(y_0-0)^2}=\left| y_0+2 \right|\] now solve this:
AlexandervonHumboldt2
  • AlexandervonHumboldt2
square both sides: \[(x_0-2)^2+y_0^2=(y_0+2)^2\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
simplify: \[x^2-2x+4+y^2=y^2+2y+4\]

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AlexandervonHumboldt2
  • AlexandervonHumboldt2
simplify: \[x^2-2x+4-4+y^2-y^2=2y\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[x^2-2x=2y\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[y=\frac{ x^2-2x }{ 2 }\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
i might have done a mistake though haven't made such question for long time
anonymous
  • anonymous
thats not any of my choices :/
AlexandervonHumboldt2
  • AlexandervonHumboldt2
what are your choises?
anonymous
  • anonymous
\[f(x)= -\frac{ 1 }{ 8 } (x-2)^{2}-2\]
anonymous
  • anonymous
\[f(x)=\frac{ 1 }{ 16 } (x-2)^{2}-2\]
anonymous
  • anonymous
and then the same as the first except positive 1/8 and same as the second but negative 1/16
AlexandervonHumboldt2
  • AlexandervonHumboldt2
OHHHHHHHHHHHHH i counted with focus (2, 0) not (2, 6) wait lemme redo it
anonymous
  • anonymous
okey
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[\sqrt{(x_0-2)^2+(y_0-6)^2}=\left| y_0+2 \right|\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[(x_0-2)^2+(y_0-6)^2=(y_0+2)^2\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[(x-2)^2+y^2-12y+36=y^2+2y+4\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[(x-2)^2+32=14y\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
woops mistake again
AlexandervonHumboldt2
  • AlexandervonHumboldt2
(x-2)^2+y^2-12y+36=y^2+4y+4
AlexandervonHumboldt2
  • AlexandervonHumboldt2
(x-2)^2+32=16y y=1/16*(x-2)^2+2 i the answer if i again didn't made a mistake somewhere
anonymous
  • anonymous
thanks! ill let you know in 1 sec
anonymous
  • anonymous
yep! it was right!! i got 100/100 on my quiz thanks!
AlexandervonHumboldt2
  • AlexandervonHumboldt2
np sorry for taking so much time
AlexandervonHumboldt2
  • AlexandervonHumboldt2
np sorry for taking so much time

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