Question

helpppp

Answer 1

@mathstudent55

Answer 2

\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] now the center is \((h,k)\) which in your case is \((-1,-3)\) so we can start with
\[\frac{(x+1)^2}{a^2}+\frac{(y+3)^2}{b^2}=1\]

Answer 3

now you need \(a\) and \(b\)
do you know how to find them?

Answer 4

no

Answer 5

ok pretty easy in this example
what is half of 6 ?

Answer 6

3

Answer 7

ok good, so \(a=3\) and therefore \(a^2=?\)

Answer 8

9

Answer 9

ok so it is \[\frac{(x+1)^2}{9}+\frac{(y+3)^2}{b^2}=1\]
now you have to find \(b^2\) which you do in a similar manner

Answer 10

what is half of 4?

Answer 11

2

Answer 12

and \(2^2=4\) so final answer is \[\frac{(x+1)^2}{9}+\frac{(y+3)^2}{4}=1\]