Write an equation for an ellipse centered at

Ответ:

golfthrash

22.01.2022

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

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