Answer :
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
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).