The foci of an ellipse are (± 5, 0) and eccentricity is -2, Find the equation of the ellipse.
Answers
EXPLANATION.
Foci of an ellipse = (±5, 0).
eccentricity = -2.
As we know that,
General formula of an ellipse.
⇒ x²/a² + y²/b² = 1.
We can write foci as,
⇒ (5,0) and (-5,0).
Using the distance formula, we get.
⇒ √(x₁ - x₂)² + (y₁ - y₂)².
⇒ √(5 - (-5))² + (0 - 0)².
⇒ √(5 + 5)².
⇒ √(10)².
⇒ √100 = 10.
Foci of an ellipse = (ae,0) and (-ae,0).
⇒ 2ae = 10.
⇒ ae = 5.
Squaring on both sides, we get.
⇒ (ae)² = (5)².
⇒ a²e² = 25.
Put the value of eccentricity = -2 in equation, we get.
⇒ a²(-2)² = 25.
⇒ a²4 = 25.
⇒ a² = 25/4.
As we know that,
Formula of eccentricity.
⇒ b² = a²(1 - e²).
⇒ b² = a² - a²e².
⇒ b² = 25/4 - 25.
⇒ b² = 25 - 100/25.
⇒ b² = -75/4.
Put the values in general equation of ellipse, we get.
⇒ x²/a² + y²/b² = 1.
⇒ x²/25/4 + y²/-75/4 = 1.
⇒ 4x²/25 - 4y²/75 = 1.
MORE INFORMATION.
Standard form of the equation of ellipse.
(1) = x²/a² + y²/b² = 1 (when a > b).
(a) = Length of major axis = A₁A₂ = 2a.
(b) = Length of minor axis = B₁B₂ = 2b.
(c) = Vertices (a,0) and (-a,0).
(d) = Directrices = x = a/e and x = -a/e.
(e) = Foci : S(ae,0) and S'(-ae,0).
(f) = Length of latus rectum = 2b²/a.
(g) = eccentricity = b² = a²(1 - e²) Or e = √1 - b²/a².
(2) = x²/a² + y²/b² = 1 (when b < a).
(a) = Length of major axis = A₁A₂ = 2b.
(b) = Length of minor axis = B₁B₂ = 2a.
(c) = Vertices (0,b) and (0,-b).
(d) = Directrices, y = ±b/e.
(e) = Foci, (0, ±be).
(f) = Length of latus rectum = 2a²/b.
(g) = Eccentricity = a² = b²(1 - e²) Or e = √1 - a²/b².