Question

Given the information below, write the equation in Standard Form.

Answer 1

Answer:

The standard form of the equation  of the ellipse is  $\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1$

Step-by-step explanation:

The standard form of the equation of an ellipse with center (h , k)  is

$\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$ , 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  $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$

- Substitute the values of h, k, a, and b in the equation

∴  $\frac{(x--3)^{2}}{2^{2}}+\frac{(y-4)^{2}}{3^{2}}=1$

∴  $\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1$

The standard form of the equation  of the ellipse is  $\frac{(x+3)^{2}}{4}+\frac{(y-4)^{2}}{9}=1$