Question

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

Answer 1

Center of the ellipse is `C(0,0) and`

foci are `S_1(0,-sqrt7) and S_2(0,sqrt7)`

Explanation:

We have ,the eqn. of ellipse is :

`x^2/9+y^2/16=1`

`Method :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` ,`"then the foci of ellipse are:"`

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

where, `c " is the distance of each focus from the center , " c > 0`

`diamondc^2`=`a^2-b^2` when , `(a > b) and c^2`=`b^2-a^2`when ,(a < b )

Comparing the given eqn.

`(x-0)^2/9+(y-0)^2/16=1`

We get ,`h=0,k=0,a^2=9 and b^2=16`

So,the center of the ellipse is =`C(h,k)=C(0,0)`

`a < b=>c^2=b^2-a^2=16-9=7=>c=sqrt7`

So, the foci of ellipse are :

`S_1(h,k-c)=S_1(0,0-sqrt7)=S_1(0,-sqrt7)`

`S_2(h,k+c)=S_2(0,0+sqrt7)=S_1(0,sqrt7)`

For second method please see next answer.

Answer 2

Center of ellipse is =`C(0,0) and`
`S_1(0,-sqrt7) and S_2(0,sqrt7)`#

Explanation:

We have ,
`x^2/9+y^2/16=1......to(1)`

`"Method : II`

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

`x^2/a^2+y^2/b^2=1 , then ,`

Center of ellipse is =`C(0,0) and`

Foci of ellipse are :

`S_1(0,-be) and S_2(0,be) ,`

`"where e is the eccentricity of ellipse"`

`e=sqrt(1-b^2/a^2) ,when, a > b`

`e=sqrt(1-a^2/b^2) ,when, a < b`

Comparing the given eqn. `(1)` we get

`a^2=9 and b^2=16=>a=3 and b=4 ,where, a < b`

`:.e=sqrt(1-a^2/b^2)=sqrt(1-9/16)=sqrt(7/16)=sqrt7/4`

So ,the foci of ellipse are :

`S_1(0,-be)=(0,-4*sqrt7/4)=>S_1(0,-sqrt7)`

`S_2(0,be)=(0,4*sqrt7/4)=>S_2(0,sqrt7)`