Question

Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

Answer 1

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:

Answer 2

square both sides:
\[(x_0-2)^2+y_0^2=(y_0+2)^2\]

Answer 3

simplify:
\[x^2-2x+4+y^2=y^2+2y+4\]

Answer 4

simplify:
\[x^2-2x+4-4+y^2-y^2=2y\]

Answer 5

\[x^2-2x=2y\]

Answer 6

\[y=\frac{ x^2-2x }{ 2 }\]

Answer 7

i might have done a mistake though

haven't made such question for long time

Answer 8

thats not any of my choices :/

Answer 9

what are your choises?

Answer 10

\[f(x)= -\frac{ 1 }{ 8 } (x-2)^{2}-2\]

Answer 11

\[f(x)=\frac{ 1 }{ 16 } (x-2)^{2}-2\]

Answer 12

and then the same as the first except positive 1/8 and same as the second but negative 1/16

Answer 13

OHHHHHHHHHHHHH i counted with focus (2, 0) not (2, 6) wait lemme redo it

Answer 14

okey

Answer 15

\[\sqrt{(x_0-2)^2+(y_0-6)^2}=\left| y_0+2 \right|\]

Answer 16

\[(x_0-2)^2+(y_0-6)^2=(y_0+2)^2\]

Answer 17

\[(x-2)^2+y^2-12y+36=y^2+2y+4\]

Answer 18

\[(x-2)^2+32=14y\]

Answer 19

woops mistake again

Answer 20

(x-2)^2+y^2-12y+36=y^2+4y+4

Answer 21

(x-2)^2+32=16y
y=1/16*(x-2)^2+2 i the answer if i again didn't made a mistake somewhere

Answer 22

thanks! ill let you know in 1 sec

Answer 23

yep! it was right!! i got 100/100 on my quiz thanks!

Answer 24

np

sorry for taking so much time

Answer 25

np

sorry for taking so much time