Question

If the ellipse given by ((x - 3)^2)/5^2 + ((y - 2)^2)/3^2 = 1 is shifted two units up, what is the translated equation?

Answer 1

\[\frac{ (x-3 ^2)}{ 5^2}+\frac{( y-2^2) }{ 5^2 }=1\]

Answer 2

@Michele_Laino

Answer 3

((x - 3)^2)/5^2 + ((y - 2)^2)/5^2 = 1

Answer 4

Thats A

Answer 5

((x - 1)^2)/5^2 + ((y - 2)^2)/3^2 = 1 Thats B

Answer 6

((x - 3)^2)/5^2 + ((y - 4)^2)/3^2 = 1 This is C

Answer 7

((x - 5)^2)/5^2 + ((y - 2)^2)/3^2 = 1 is D

Answer 8

if we perform this traslation:
x-3=X
y-2=Y
where X and Y are the new coordinates, then we can rewrite your ellipse as below:
\[\frac{{{X^2}}}{{25}} + \frac{{{Y^2}}}{9} = 1\]

Answer 9

(x-3)^2/25+y-2/9=1?

Answer 10

the origin of the new coordinate system XOY is located at this point (3,2)

Answer 11

please wait it is the first step only

Answer 12

ok

Answer 13

now if I perform the traslation indicated by your problem, namely:
X'=X
Y?=Y-1
where X' and Y' are new coordinates, then we have:

In that new reference system X?OY? your ellipse is:
\[\frac{{X{'^2}}}{{25}} + \frac{{Y' + {1^2}}}{9} = 1\]

Answer 14

now we have:
y-2=Y'+2
or
Y'=y-4
and
x-3=X'
so substituting, we get:
\[\Large \frac{{{{\left( {x - 3} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 4} \right)}^2}}}{9} = 1\]
sorr y the new coordinates, are:
X'=X
and
Y?=Y-2

Answer 15

oops..Y'=Y-2

Answer 16

ok