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

Sagot :

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:

Center of the ellipse = (0, 0)

vertices are (±√8, 0) and (0, ±√5)

Focus of the ellipse =  (±√3, 0).

Step-by-step explanation:

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

We will rewrite this equation in the vertex form

[tex]\frac{5x^{2}+8y^{2}}{40}=\frac{40}{40}[/tex]

⇒[tex]\frac{x^{2}}{8}+\frac{y^{2}}{5}=1[/tex]

⇒[tex]\frac{(x-0)^{2}}{8}+\frac{(y-0)^{2}}{5}=1[/tex]

This equation is in the form of

⇒[tex]\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1[/tex]

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 =[tex]\sqrt{a^{2}-b^{2}}[/tex]

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).