The co-ordinates of foci of the ellipse $16 x^{2}+9 y^{2}=144$ are
The co-ordinates of foci of the ellipse $16 x^{2}+9 y^{2}=144$ are
$(\pm 7,0)$
$(0, \pm \sqrt{7})$
$(\pm \sqrt{7}, 0)$
$(0, \pm 7)$
Solution
Center is $(0,0)$, focii are $(0, \sqrt{7})$ and $(0,-\sqrt{7})$, major axis is along $y$-axis and is 8 , and minor axis is along $x$-axis and is 6 .
Explanation:
Note that equation $16 x^{2}+9 y^{2}=144$ by dividing each term by 144 can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
As the standard equation is of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, center is $(0,0)$.
In this equation major axis is along $y$-axis and is $2 b=8$, and minor axis is along $x$ axis and is $2 a=6$.
As $a=3$ and $b=4$, focii are given by $\pm \sqrt{4^{2}-3^{2}}=\pm \sqrt{7}$ i.e. $(0, \sqrt{7})$ and $(0,-\sqrt{7})$