Respuesta :
Answer:
The standard form of the equation of the ellipse is [tex]\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1[/tex]
Step-by-step explanation:
The standard form of the equation of an ellipse with center (h , k) is
[tex]\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1[/tex] , where
- The coordinates of the vertices are (h , k ± a)
- The coordinates of the co-vertices are (h ± b , k)
- The coordinates of the foci are (h , k ± c), where c² = a² - b²
∵ The vertices of the ellipse are (-3 , 7), (-3 , 1)
∴ h = -3
∴ k + a = 7 ⇒ (1)
∴ k - a = 1 ⇒ (2)
- Add (1) and (2) to find k
∴ 2k = 8
- Divide both sides by 2
∴ k = 4
- Substitute the value of k in (1) or (2) to find a
∵ 4 + a = 7
- Subtract 4 from both sides
∴ a = 3
∵ The co-vertices of the ellipse are (-5 , 4), (-1 , 4)
∴ k = 4
∴ h - b = -5 ⇒ (1)
∴ h + b = -1 ⇒ (2)
∵ h = -3
- Substitute the value of h in (1) or (2) to find b
∴ -3 + b = -1
- Add 3 to both sides
∴ b = 2
∵ The standard form of the equation of the ellipse is [tex]\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1[/tex]
- Substitute the values of h, k, a, and b in the equation
∴ [tex]\frac{(x--3)^{2}}{2^{2}}+\frac{(y-4)^{2}}{3^{2}}=1[/tex]
∴ [tex]\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1[/tex]
The standard form of the equation of the ellipse is [tex]\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1[/tex]