Question

Find the center, vertices, and foci of the ellipse with equation 2x2 + 7y2 = 14.

Center: (0, 0); Vertices: (-7, 0), (7, 0); Foci: Ordered pair negative 3 square root 5 comma 0 and ordered pair 3 square root 5 comma 0
Center: (0, 0); Vertices: Ordered pair negative square root 7 comma 0 and ordered pair square root 7 comma 0; Foci: Ordered pair negative square root 5 comma 0 and ordered pair square root 5 comma 0
Center: (0, 0); Vertices: Ordered pair 0 comma negative square root 7 and ordered pair 0 comma square root 7; Foci: Ordered pair 0 comma negative square root 5 and ordered pair 0 comma square root 5
Center: (0, 0); Vertices: (0, -7), (0, 7); Foci:Ordered pair 0 comma negative 3 square root 5 and ordered pair 0 comma 3 square root 5

Answer 1

Answer:

Center: (0 , 0); Vertices: ($-\sqrt{7}$ , 0) and ($\sqrt{7}$ , 0); Foci: ($-\sqrt{5}$ , 0) and ($\sqrt{5}$ , 0) ⇒ 2nd answer

Step-by-step explanation:

The standard form of the equation of an ellipse with center (0 , 0 )  and major axis parallel to the x-axis is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , a > b  , where

• The coordinates of the vertices are (± a , 0)
• The coordinates of the foci are (± c , 0), and c ² = a² - b²  

∵ The equation of the ellipse is 2x² + 7y² = 14

- Divide both sides by 14 to make the right hand side = 1

∴ $\frac{2x^{2}}{14}+\frac{7y^{2}}{14}=\frac{14}{14}$

- Simplify the fractions

∴  $\frac{x^{2}}{7}+\frac{y^{2}}{2}=1$

Compare it with the form of the ellipse above

∴ The center of the ellipse is (0 , 0)

∴ a² = 7

- Take √  for both sides

∴ a = ± $\sqrt{7}$

∴ b² = 2

- Take √  for both sides

∴ b = ± $\sqrt{2}$

∵ The vertices of it are (a , 0) and (-a , 0)

∴ Its vertices are ($\sqrt{7}$ , 0) and ($-\sqrt{7}$ , 0)

∵ c² = a² - b²

∵ a² = 7 and b² = 2

∴ c² = 7 - 2

∴ c² = 5

- Take √  for both sides

∴ c = ± $\sqrt{5}$

∵ The foci of it are (c , 0) and (-c , 0)

∴ Its foci are ($\sqrt{5}$ , 0) and ($-\sqrt{5}$ , 0)