Question

HELP. How long is the minor axis for the ellipse shown below?

Answer 1

Given:

The equation of ellipse is

$\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1$

To find:

The length of the minor axis.

Solution:

The standard form of an ellipse is

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$      ...(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,

$\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1$      ...(ii)

On comparing (i) and (ii), we get

$b^2=16$

Taking square root on both sides.

$b=\pm 4$

Consider only positive value of b because length cannot be negative.

$b=4$

Now,

Length of minor axis = $2b$

                                  = $2(4)$

                                  = $8$

So, the length of minor axis is 8 units.

Therefore, the correct option is B.