find the encentricity,the foci and vertex and the equation of directrix and asymptote to the equation give the ellipse below
25x^2+4y^2-50x-16y-59=0
25x^2+4y^2-50x-16y-59=0
25(x^2 - 50x + ...) + 4(y^2 - 4y + ..) = 59
25(x^2 - 2x + 1) + 4(y^2 - 4y + 4) = 59 + 25 + 16
25(x-1)^2 + 4(y-2)^2 = 100
(x-1)^2 /4 + (y-2)^2 /25 = 1
Now just use all the little formulas you must have to state the properties of this nice ellipse
To find the eccentricity, foci, vertex, equation of the directrix, and asymptotes of the ellipse given by the equation: 25x^2 + 4y^2 - 50x - 16y - 59 = 0, follow these steps:
Step 1: Rearrange the equation to standard form
Start by grouping the x and y terms together and moving the constant term to the other side of the equation:
25x^2 - 50x + 4y^2 - 16y = 59
Divide the entire equation by the constant term on the right side to obtain:
x^2 - 2x + (4/25)y^2 - (4/25)y = 59/25
Step 2: Complete the square for x and y terms
Now, complete the square for both x and y terms by adding and subtracting appropriate constants.
For x terms, take half of the coefficient of x (-2/2 = -1) and square it to get 1. Add and subtract 1 on the x side:
x^2 - 2x + 1 + (4/25)y^2 - (4/25)y = 59/25 + 1 - 1
For y terms, take half of the coefficient of y (-4/25 / 2 = -2/25) and square it to get 4/625. Add and subtract 4/625 on the y side:
x^2 - 2x + 1 + (4/25)y^2 - (4/25)y + 4/625 - 4/625 = 59/25 + 1 - 4/625 + 4/625
Simplify:
(x - 1)^2 + (y - 2/25)^2 = 1561/625
Step 3: Identify the constants for the standard form equation
From the completed square form, we can see that:
- The x-coordinate of the vertex is h = 1.
- The y-coordinate of the vertex is k = 2/25.
Step 4: Calculate the eccentricity
The eccentricity (ε) of an ellipse can be found using the equation: ε = √(1 - b^2/a^2), where a is the semi-major axis and b is the semi-minor axis.
In the given standard form equation, we already have the values of a and b:
a^2 = (1561/625) / 1 = 1561/625
b^2 = (1561/625) / (4/25) = 9/625
Now, substitute these values into the eccentricity formula:
ε = √(1 - (9/625) / (1561/625))
ε = √(1 - 9/1561)
ε = √(1552/1561)
ε ≈ 0.993
Step 5: Identify the foci
The foci distances (c) can be found using the equation: c = √(a^2 - b^2)
Using the values of a^2 and b^2 obtained earlier, calculate c:
c = √(1561/625 - 9/625)
c = √(1552/625)
c ≈ √2.4832
The foci are located at coordinates (h ± c, k). Substituting the values, we get:
Foci: (1 ± √2.4832, 2/25)
Step 6: Determine the equation of the directrix
The equation of the directrix is given by: x = h ± a/ε for horizontal ellipses, and y = k ± a/ε for vertical ellipses.
Using the values of h, a, and ε, substitute into the equation to find the directrix:
Directrix: x = 1 ± (1561/625) / (0.993)
Step 7: Find the asymptotes
The slopes of the asymptotes can be calculated using the formula: m = ± b / a.
Given the values of b and a, calculate the slopes:
m = ± (3/25) / √(1561/625)
The equations of the asymptotes are given by: y - k = m(x - h) for horizontal ellipses or x - h = m(y - k) for vertical ellipses.
Substituting the values of h, k, and m, we get:
Asymptotes: y - 2/25 = ± [(3/25) / √(1561/625)] (x - 1)
Now, we have:
- Eccentricity (ε) ≈ 0.993
- Foci: (1 ± √2.4832, 2/25)
- Vertex: (1, 2/25)
- Equation of Directrix: x = 1 ± (1561/625) / (0.993)
- Equations of Asymptotes: y - 2/25 = ± [(3/25) / √(1561/625)] (x - 1)