Question

The vertices of the ellipse $\frac{(x+1{)}^2}{25}+\frac{(y-3{)}^2}{16}$ = 1 is

• A. (3,4) and (-6,3)
• B. (4,3) and (-6,3)
• C. (5,3) and (-7,3)
• D. (6,3) and (-8,3)

Answer 1

The correct option is B

(4,3) and (-6,3)

$\frac{(x+1{)}^2}{25}+\frac{(y-3{)}^2}{16}$ = 1, a=5, b=4,centre (h,k)=(-1,3)

Vertices = (h$\pm$a,k)=(-1$\pm$5,3)=(4,3) and (-6,3)