Question

Derive the equation of the parabola with a focus at (-7, 5) and a directrix of y = -11.

Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = -1

Answer 1

I got the answer to the first one is \[f(x)=\frac{ 1 }{ 32 }(x+7)^2-3\]

Answer 2

and I know the second one has \[\frac{ 1 }{ 12 } \] and \[(x + 5)^2 + 2\] or \[(x - 5)^2 + 2\]

Answer 3

post the work of one of the question first or 2nd

Answer 4

@Nnesha (x - x)^2 + (y - -11)^2 = (x - -7)^2 + (y - 5)^2
y^2 + 22y + 121 = (x + 7)^2 + y^2 -10y + 25
32 y = (x + 7)^2 -96
y = (1/32)(x + 7)^2 - 3

Answer 5

for the first one ^^

Answer 6

as for the second one I know that because all the answers have those thing(s) in them

Answer 7

@Nnesha please help me with the second one

Answer 8

2nd would be the same
and the equation is to find distance between focus and point on the parabola
and distance between directrix and point \[\rm \sqrt{(x-a)^2+(y-b)^2}=\left| y-c \right|\]
square both sides \[\rm (x-a)^2+(y-b)^2=(y-c)^2\]
where (a,b) is the focus point and c is diretrix

Answer 9

plug in (-5,5) for (a,b) and -1 for c
lets see which one u get

Answer 10

@Nnesha is it 1/12 (x+5)^2 + 2

Answer 11

why not (x−5)^2+2? it would be easy for me to check if u post the work :=))