Question

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

Answer 1

Answer:

C is the correct answer

Step-by-step explanation:

Answer 2

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.