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

A.) (6, –3) and (6, –11)
B.) 6, –5) and (6, –9)
C.) (10, –7) and (2, –7)
D.)(8, –7) and (4, –7)

User Genu
by
9.1k points

2 Answers

4 votes

Answer:

C is the correct answer

Explanation:

User Shartte
by
8.4k points
7 votes

Answer:

The correct option is C.

Explanation:

The given equation is


((x-6)^2)/(16)+((y+7)^2)/(4)= 1

It can be rewritten as


((x-6)^2)/(4^2)+((y+7)^2)/(2^2)= 1 .....(1)

The standard form of an ellipse is


((x-h)^2)/(a^2)+((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.

User Rtist
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9.5k points

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