Answer
Different centres (the endpoints, the foci are translated accordingly)
Work Step by Step
We are given the two ellipses:
$\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$
$\dfrac{(x-1)^2}{25}+\dfrac{(y-1)^2}{16}=1$
The standard equation for each ellipse is:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,a,b,c$ for the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$:
$h=0$
$k=0$
$a^2=25\Rightarrow a=\sqrt{25}=5$
$b^2=16\Rightarrow b=\sqrt{16}=4$
$c^2=a^2-b^2$
$c^2=25-16$
$c^2=9$
$c=\sqrt 9=3$
The centre of the ellipse is $C_1(0,0)$, the length of the major axis is $2a=2(5)=10$, the length of the minor axis is $2b=2(4)=8$.
Determine $h,k,a,b,c$ for the ellipse $\dfrac{(x-1)^2}{25}+\dfrac{(y-1)^2}{16}=1$:
$h=1$
$k=1$
$a^2=25\Rightarrow a=\sqrt{25}=5$
$b^2=16\Rightarrow b=\sqrt{16}=4$
$c^2=a^2-b^2$
$c^2=25-16$
$c^2=9$
$c=\sqrt 9=3$
The centre of the ellipse is $C_2(1,1)$, the length of the major axis is $2a=2(5)=10$, the length of the minor axis is $2b=2(4)=8$.
The two ellipses have different centres (the endpoints, the foci are translated accordingly).
