Question

Can anyone help me PLEASE ;
find the center, vertices, and foci of this ellipse.
1.(x+6)^2/12+(y-4)^2/16=1

4.x^2+4y^2+8x-48=0

Answer 1

@rsadhvika thanks alot. what formula do i use for number 4?

Answer 2

you figured out first one ?

Answer 3

working on it now . I'm trying !

Answer 4

careful first one if of form :-

\(\large \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)

Answer 5

so you need to use formulas from last column in the given table ok

Answer 6

ok

Answer 7

did u get, why it is of that form ? :)

Answer 8

not yet !

Answer 9

well it is of that form cuz, 12 < 16
so \(b\) goes under x-thing

Answer 10

ok , i got lost on the whole assignment. I guess I have been doing it work . :(

Answer 11

First step is you need to figure out, which form the given ellipse equation is.

Answer 12

once you're clear about that,
rest is bit easy

Answer 13

1.(x+6)^2/12+(y-4)^2/16=1

\(\large \frac{(x+6)^2}{12} + \frac{(y-4)^2}{16} = 1\)


\(\large \frac{(x+6)^2}{(\sqrt{12})^2} + \frac{(y-4)^2}{4^2} = 1\)

Answer 14

since 12 under x, is less than 16
you need to use formulas from last column.

a = 4
b = \(\sqrt{12}\)
h = -6
k = 4

Answer 15

once u digest above, let me knw

Answer 16

ok.

Answer 17

i think i have it .