Question

2(x-4)^2+2y^2=8 please help me find radius and center

Answer 1

divide all of it by 2 first

Answer 2

both sides?

Answer 3

of course

Answer 4

how do you divide 2 by (x-4)^2?

Answer 5

factor 2 out of (x-4)^2 and y^2 so you can divide by 2 on both sides

Answer 6

oh so dived 2 by x^2-8x+16?

Answer 7

dont expand (x-4)^2, leave it alone

Answer 8

would x^2 , the 1st term still be x^2? BECAUSE I CAN DIVIDE THE REST BUT THAT 1ST TERM

Answer 9

sorry for caps

Answer 10

ohhhh

Answer 11

got it?

Answer 12

no :(

Answer 13

so would it look like : (x-4)^2+2y^2/2=8 /2

Answer 14

yes

Answer 15

now divide it out

Answer 16

can you give me an example? please , on how to divide (x-4)^2 by 2? byt giving me an example

Answer 17

you already have
in the problem its 2(x-4)^2

Answer 18

so it would look like 2(x-4)^2+y^2=4?

Answer 19

it would look like (x-4)^2 + y^2 = 4

Answer 20

ohhh ok thanks the can you help me get it to standard form and find center and radius and intercepts?

Answer 21

you can get radius + center from this

Answer 22

wait , so the center would be (4,1) and radius is 4?

Answer 23

radius is square root of number on right
center is (4, 0) because y^2 is by itself, it doesnt look like (y - 1)^2

Answer 24

so y^2 is not same as 1y^2? its 0y^2?

Answer 25

also I got the intercepts as (-3,0)(0,-8)

Answer 26

y^2 is equal to 1y^2 @bunnybee but you have (y-0)^2 so the second coordinate is 0 not y...remember that in looking for the coordinates of the center..you look at the second term of the binomial not the coefficient of the variable ^_^ make any sense?

Answer 27

yes thank you!