Question

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

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

Answer 1

Answer: (D)

$( - 1,2),( - 1, - 6)$

Answer 2

Given ellipse is $\frac{(x + 1)^{2}}{9} + \frac{(y + 2)^{2}}{25} = 1$ i.e. $\frac{X^{2}}{9} + \frac{Y^{2}}{25} = 1$, where $X = x + 1$and $Y = y + 2$

Here $a^{2} = 25,b^{2} = 9$ [Type : $\frac{X^{2}}{b^{2}} + \frac{Y^{2}}{a^{2}} = 1\rbrack$

Eccentricity is given by $e^{2} = \frac{a^{2} - b^{2}}{a^{2}} = \frac{25 - 9}{25} = \frac{16}{25}$, $\therefore e = \frac{4}{5}$

Foci are given by $Y = \pm ae = \pm 5\left( \frac{4}{5} \right) = \pm 4$

$X = 0$⇒ $y + 2 = \pm 4$⇒$y = - 2 \pm 4 = - 6$ or 2

$x + 1 = 0$ ⇒ $x = - 1$. Hence foci are (–1, –6) or (–1, 2).