Question

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)

Answer 1

C is the correct answer

Explanation:

Answer 2

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.