Answer:
The correct option is C.
Step-by-step explanation:
The given equation is
[tex]\frac{(x-6)^2}{16}+\frac{(y+7)^2}{4}= 1[/tex]
It can be rewritten as
[tex]\frac{(x-6)^2}{4^2}+\frac{(y+7)^2}{2^2}= 1[/tex] .....(1)
The standard form of an ellipse is
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= 1[/tex] ....(2)
Where (h,k) is center of the ellipse.
If a>b, then the vertices of the ellipse are [tex](h\pm a, k)[/tex].
From (1) and (2) we get
[tex]h=6,k=-7,a=4,b=2[/tex]
Since a>b, therefore the vertices of the ellipse are
[tex](h+a, k)=(6+4,-7)\Rightarrow (10,-7)[/tex]
[tex](h-a, k)=(6-4,-7)\Rightarrow (2,-7)[/tex]
The vertices of the given ellipse are (10, –7) and (2, –7). Therefore the correct option is C.
Answer:
C is the correct answer
Step-by-step explanation:
