Question

The foci of the ellipse $25(x+1{)}^2+9(y+2{)}^2=225$ are at

• A. $(-1,2)$ and $(-1,-6)$
• B. $(-1,2)$ and $(6,1)$
• C. $(1,-2)$ and $(1,-6)$
• D. $(-1,-2)$ and $(1,6)$

Answer 1

The correct option is A $(-1,2)$ and $(-1,-6)$

Given that

$25(x+1{)}^2+9(y+2{)}^2=225$

Dividing throughout by 225, we get

$\frac{x+1{)}^2}{9}+\frac{(y+2{)}^2}{25}=1$

Assume $X=x+1$ and $Y=y+2$ just for the simplicity

Now the equation is,

$\frac{X^2}{9}+\frac{Y^2}{25}=1$

Here, we can see that the denominator of $Y^2$ is greater, which means that the major axis of the ellipse lies along the y-axis.

So, the equation of the ellipse is of the form

$\frac{X^2}{b^2}+\frac{Y^2}{a^2}=1$

where, $b=3$ and $a=5$

Therefore,

$c=\sqrt{a^2-b^2}=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=\pm 4$

Hence,

The foci are $(X=0,Y=\pm c)$ i.e. $(x+1=0,y+2=\pm 4)$

i.e. $(-1,2)$ and $(-1,-6)$