Question

I need help like RN!!!
what are the foci of the ellipse? graph the ellipse. 18x^2+36y^=648

Answer 1

Hi,


`18x^2+36y^2=648\\ \Rightarrow ( (x)/(6))^2+( (y)/(3* √(2) ) )^2=1\\ a=6,\ b=3* √(2)\\ c= √(a^2-b^2) = √(36-18) =3* √(2)\\`
Center is (0,0)
Foci are (-3√2,0) and (3√2,0)

Answer 2

The equation of ellipse is given by:

`(x^2)/(a^2)+(y^2)/(b^2) =1` ....[1]

where,

a and b are the semi major axis and semi minor axis.

Foci of ellipse =
`(\pm c, 0)`

where,
`c = √(a^2-b^2)`

As per the statement:

Given the ellipse is:

`18x^2+36y^2 = 648`

Divide both sides by 648 we have;

`(x^2)/(36)+(y^2)/(18) =1`

On comparing with [1] we have;

`a^2 =36` and
`b^2=18`

First find c:

`c = √(36-18)=√(18)=3√(2)`

Foci of the ellipse are:

`(\pm 3√(2), 0)`

Therefore, the foci of the ellipse are,
`(3√(2), 0)` and
`(-3√(2) , 0)`