Question

Write an equation for an ellipse centered at the origin, which has foci at (\pm8,0)(±8,0)left parenthesis, plus minus, 8, comma, 0, right parenthesis and vertices at (\pm17,0)(±17,0)left parenthesis, plus minus, 17, comma, 0, right parenthesis

Answer 1

Answer:

The equation of the ellipse is $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$

Step-by-step explanation:

The standard form of the equation of an ellipse with center (0 , 0) is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where

• The coordinates of the vertices are (± a , 0)  

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

∵ The ellipse is centered at the origin

∴ The center of it is (0 , 0)

∵ It has foci at (± 8 , 0)

- The coordinates of the foci are (± c , 0)

∴ c = ±8

∵ It has Vertices at (± 17 , 0)

- The coordinates of the vertices are (± a , 0)

∴ a = ±17

∵ c² = a² - b²

- Substitute the values of c and a to find b

∴ (8)² = (17)² - b²

∴ 64 = 289 - b²

- Add b² to both sides

∴ b² + 64 = 289

- Subtract 64 from both sides

∴ b² = 225

- Take √  for both sides

∴ b = ± 15

∵ The equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

- Substitute the values of a and b in it

∴ $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$ ⇒ $\frac{x^{2}}{289}+\frac{y^{2}}{225}=1$  

∴ The equation of the ellipse is $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$