Question

Find the center, vertices, and foci of the ellipse with equation 4x2 + 9y2 = 36. Center: (0, 0); Vertices: (-3, 0), (3, 0); Foci: Ordered pair negative square root 5 comma 0 and ordered pair square root 5 comma 0 Center: (0, 0); Vertices: (-9, 0), (9, 0); Foci: Ordered pair negative square root 65 comma 0 and ordered pair square root 65 comma 0 Center: (0, 0); Vertices: (0, -3), (0, -3); Foci: Ordered pair 0 comma negative square root 5 and ordered pair 0 comma square root 5 Center: (0, 0); Vertices: (0, -9), (0, 9); Foci: Ordered pair 0 comma negative square root 65 and ordered pair 0 comma square root 65

Answer 1

Option A.

Explanation:

The given equation of ellipse is

`4x^2+9y^2=36`

Divide both sides by 36.

`(4x^2)/(36)+(9y^2)/(36)=1`

`(x^2)/(9)+(y^2)/(4)=1`

`(x^2)/(3^2)+(y^2)/(2^2)=1` ...(1)

The standard form of an ellipse is

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1` ...(2)

where, (h,k) is center, (h±a,k) are vertices and (h±c,k) are foci.

On comparing (1) and (2), we get

`h=0,k=0,a=3,b=2`

Now,

Center
`=(h,k)=(0,0)`

Vertices
`=(h\pm a,k)=(0\pm 3,0)=(3,0),(-3,0)`

We know that

`c=√(a^2-b^2)=√(3^2-2^2)=√(5)`

Foci
`=(h\pm c,k)=(0\pm √(5),0)=(√(5),0),(-√(5),0)`

Therefore, the correct option is A.