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dangina [55]
2 years ago
12

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

Mathematics
1 answer:
kifflom [539]2 years ago
8 0

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.

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