# What are the center and foci of the ellipse described by x^2/9 + y^2/16 =1?

Aug 10, 2018

Center of the ellipse is $C \left(0 , 0\right) \mathmr{and}$

foci are ${S}_{1} \left(0 , - \sqrt{7}\right) \mathmr{and} {S}_{2} \left(0 , \sqrt{7}\right)$

#### Explanation:

We have ,the eqn. of ellipse is :

${x}^{2} / 9 + {y}^{2} / 16 = 1$

$M e t h o d : I$

If we take standard eqn. of ellipse with center color(red)(C (h,k) ,as

color(red)((x-h)^2/a^2+(y-k)^2/b^2=1 ,$\text{then the foci of ellipse are:}$

color(red)(S_1(h,k-c) and S_2(h,k+c),

where, $c \text{ is the distance of each focus from the center , } c > 0$

$\diamond {c}^{2}$=${a}^{2} - {b}^{2}$ when , $\left(a > b\right) \mathmr{and} {c}^{2}$=${b}^{2} - {a}^{2}$when ,(a < b )

Comparing the given eqn.

${\left(x - 0\right)}^{2} / 9 + {\left(y - 0\right)}^{2} / 16 = 1$

We get ,$h = 0 , k = 0 , {a}^{2} = 9 \mathmr{and} {b}^{2} = 16$

So,the center of the ellipse is =$C \left(h , k\right) = C \left(0 , 0\right)$

$a < b \implies {c}^{2} = {b}^{2} - {a}^{2} = 16 - 9 = 7 \implies c = \sqrt{7}$

So, the foci of ellipse are :

${S}_{1} \left(h , k - c\right) = {S}_{1} \left(0 , 0 - \sqrt{7}\right) = {S}_{1} \left(0 , - \sqrt{7}\right)$

${S}_{2} \left(h , k + c\right) = {S}_{2} \left(0 , 0 + \sqrt{7}\right) = {S}_{1} \left(0 , \sqrt{7}\right)$

Aug 10, 2018

Center of ellipse is =$C \left(0 , 0\right) \mathmr{and}$
${S}_{1} \left(0 , - \sqrt{7}\right) \mathmr{and} {S}_{2} \left(0 , \sqrt{7}\right)$

#### Explanation:

We have ,
${x}^{2} / 9 + {y}^{2} / 16 = 1. \ldots . . \to \left(1\right)$

"Method : II#

If we take , the standard eqn of ellipse with center at origin ,as

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1 , t h e n ,$

Center of ellipse is =$C \left(0 , 0\right) \mathmr{and}$

Foci of ellipse are :

${S}_{1} \left(0 , - b e\right) \mathmr{and} {S}_{2} \left(0 , b e\right) ,$

$\text{where e is the eccentricity of ellipse}$

$e = \sqrt{1 - {b}^{2} / {a}^{2}} , w h e n , a > b$

$e = \sqrt{1 - {a}^{2} / {b}^{2}} , w h e n , a < b$

Comparing the given eqn. $\left(1\right)$ we get

${a}^{2} = 9 \mathmr{and} {b}^{2} = 16 \implies a = 3 \mathmr{and} b = 4 , w h e r e , a < b$

$\therefore e = \sqrt{1 - {a}^{2} / {b}^{2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

So ,the foci of ellipse are :

${S}_{1} \left(0 , - b e\right) = \left(0 , - 4 \cdot \frac{\sqrt{7}}{4}\right) \implies {S}_{1} \left(0 , - \sqrt{7}\right)$

${S}_{2} \left(0 , b e\right) = \left(0 , 4 \cdot \frac{\sqrt{7}}{4}\right) \implies {S}_{2} \left(0 , \sqrt{7}\right)$