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VikaD [51]
2 years ago
12

The equation (x-6)^2/16 + (y+7)^2/4 = 1 represents an ellipse. What are the vertices of the ellipse?

Mathematics
2 answers:
pentagon [3]2 years ago
6 0

Answer:

C is the correct answer

Step-by-step explanation:

topjm [15]2 years ago
5 0

Answer:

The correct option is C.

Step-by-step explanation:

The given equation is

\frac{(x-6)^2}{16}+\frac{(y+7)^2}{4}= 1

It can be rewritten as

\frac{(x-6)^2}{4^2}+\frac{(y+7)^2}{2^2}= 1         .....(1)

The standard form of an ellipse is

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

Where (h,k) is center of the ellipse.

If a>b, then the vertices of the ellipse are (h\pm a, k).

From (1) and (2) we get

h=6,k=-7,a=4,b=2

Since a>b, therefore the vertices of the ellipse are

(h+a, k)=(6+4,-7)\Rightarrow (10,-7)

(h-a, k)=(6-4,-7)\Rightarrow (2,-7)

The vertices of the given ellipse are (10, –7) and (2, –7). Therefore the correct option is C.

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