Given:
The equation of ellipse is
[tex]\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1[/tex]
To find:
The length of the minor axis.
Solution:
The standard form of an ellipse is
[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex] ...(i)
where, (h,k) is center, if a>b, then 2a is length of major axis and 2b is length of minor axis.
We have,
[tex]\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]b^2=16[/tex]
Taking square root on both sides.
[tex]b=\pm 4[/tex]
Consider only positive value of b because length cannot be negative.
[tex]b=4[/tex]
Now,
Length of minor axis = [tex]2b[/tex]
= [tex]2(4)[/tex]
= [tex]8[/tex]
So, the length of minor axis is 8 units.
Therefore, the correct option is B.
