Question

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

• A. $(-2,7),(-2,-3)$
• B. $(4,3),(-6,3)$
• C. $(5,3),(-7,3)$
• D. $(6,3),(-8,3)$

Answer 1

Answer: (A)

$(-2,7),(-2,-3)$

We know vertices of standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are

$(x,y)=(\pm a,0)$ if $a>b$ and $(x,y)=(0,\pm b)$ if $b>a$

Hence vertices of ellipse $\frac{(x+2{)}^2}{16}+\frac{(y-2{)}^2}{25}=1$ are

$(x+2,y-2)=(0,\pm 5)$

$\Rightarrow (x,y)=(-2,2\pm 5)$

Vertices $(-2,2\pm 5)=(-2,7),(-2,-3)$

Hence option 'A' is correct choice.