Question

4x^2+9y^2-8x-18y-23=0 find the foci

Answer 1

do a bit of a massage to the equation

Answer 2

\[\frac{ (x-1)^2 }{ 9 }+\frac{ (y-1)^2 }{ 4 }\]

Answer 3

hmm so it is ellipse

Answer 4

It is

Answer 5

I hate these so much

Answer 6

(1,1) here is the center of this ellipse
the distance from the two foci's are equal to 2a

Answer 7

too bad i gotta go for dinner

Answer 8

now I am sht out of luck

Answer 9

it's a same question right
fo find foci first you need c

Answer 10

to*

Answer 11

It is pose to be (x1,y1),(x2,y2)

Answer 12

written in that form :(

Answer 13

open study lagging

Answer 14

\[\frac{ (x-1)^2 }{ 9 }+\frac{ (y-1)^2 }{ 4 }\]
a^2 = 9
b^2 = 4
use this formula \[\huge\rm c^2 = a^2 - b^2\]
solve for c
and bec you already know it's horizontal
that's why you should add c into x coordinate
\[\huge\rm (h \pm c, k)\]

Answer 15

c^2=65

Answer 16

\[\frac{ (x-1)^2 }{ 9 }+\frac{ (y-1)^2 }{ 4 }= 1\]
don't forget one

Answer 17

hmm how did you get 65 ??

Answer 18

nvm

Answer 19

equation
so a^2 = 9
and b^2 = 4