Question

Find the center, vertices, and foci of the ellipse with equation 5x2 + 8y2 = 40.

Answer 1

for an ellipse x²/a² + y²/a² = 1
vertices are -a,0 a, 0 0,b 0, -b focus : √(a²-b²) , 0
center is origin 0,0

given ellipse : divide by 40 on both sides

x² / 8 + y²/5 = 1
So a = √8 = 2√2 b = √5
vertices are -2√2,0 2√2,0 0,√5 0,-√5
focii = √3, 0 -√3, 0

Answer 2

Center of the ellipse = (0, 0)

vertices are (±√8, 0) and (0, ±√5)

Focus of the ellipse = (±√3, 0).

Explanation:

Equation of an ellipse is given as 5x² + 8y² = 40

We will rewrite this equation in the vertex form

`(5x^(2)+8y^(2))/(40)=(40)/(40)`

⇒
`(x^(2))/(8)+(y^(2))/(5)=1`

⇒
`((x-0)^(2))/(8)+((y-0)^(2))/(5)=1`

This equation is in the form of

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

Then Center of the ellipse is (h, k) and major vertices will be (h±a, k) with minor vertices will be (h, k±b)

and focus is (h±c, k) where c =
`\sqrt{a^(2)-b^(2)}`

Now we put the values h = 0 and k = 0

Center of this ellipse will be (0, 0)

Vertices of the ellipse will be

Major vertices = (0±√8, 0) = (±√8, 0)

Minor vertices = (0, 0±√5) = (0, ±√5)

Now Focus of the ellipse = (0±c, 0)

where c = √(a² - b²) = √(8-5) = √3

Now focus is (±√3, 0).