Give the equation of the hyperbola is 9x^2-4y^2=36.find
(i)center and vertices
(ii)foci and eccentricity
(iii)equation of assymptotes
Rewrite in standard form as:
(x/2)^2 - (y/3)^2 = 1
(i) Center is (0,0)
The vertices (lowest |x| value) are where y = 0, which would be (-2,0) amd (+2,0)
(ii) Foci are at
(0,-c) and (0,c), where
c = sqrt(2^2 + 3^2) = sqrt13 = 3.606
Eccentricity is defined as
e = c/a
In this case, a = 2, so e = c/2 = 1.803
(iii) Asymptotes are
y = (3/2)x and y = -(3/2)x
To find the equation of a hyperbola given in the form 9x^2 - 4y^2 = 36, we can follow these steps:
Step 1: Divide the equation by the constant on the right side to make the right side equal to 1:
x^2/(4/9) - y^2/9 = 36/36
Simplifying, we get:
x^2/(2/3) - y^2/9 = 1
Step 2: Identify the constants in the equation:
The constant under x^2 is 2/3, and the constant under y^2 is 9.
Step 3: Determine the value of a^2 and b^2:
The value of a^2 is the denominator of the constant under x^2, so a^2 = 2/3.
The value of b^2 is the denominator of the constant under y^2, so b^2 = 9.
Step 4: Calculate a, b, and c:
Given that a^2 = 2/3, we can find a by taking the square root of 2/3:
a = sqrt(2/3)
Given that b^2 = 9, we can find b by taking the square root of 9:
b = sqrt(9) = 3
The value of c can be found using the relationship a^2 + b^2 = c^2:
c^2 = a^2 + b^2
c^2 = 2/3 + 9
c^2 = 2/3 + 27/3
c^2 = 29/3
Taking the square root of 29/3:
c = sqrt(29/3)
Step 5: Determine the center of the hyperbola:
The center of the hyperbola is represented by the coordinates (h, k) which can be written as (0, 0) since the equation is in the standard form.
(i) Center and Vertices:
The center of the hyperbola is at the point (0, 0).
The distance from the center to each vertex along the x-axis is equal to a, so the vertices are V1 = (a, 0) and V2 = (-a, 0). Therefore, the vertices are V1 = (sqrt(2/3), 0) and V2 = (-sqrt(2/3), 0).
(ii) Foci and Eccentricity:
The distance from the center to each focus along the x-axis is equal to c, so the foci are F1 = (c, 0) and F2 = (-c, 0). Therefore, the foci are F1 = (sqrt(29/3), 0) and F2 = (-sqrt(29/3), 0).
The eccentricity can be calculated using the formula e = c/a. Therefore, the eccentricity is e = (sqrt(29/3))/sqrt(2/3).
(iii) Equation of Asymptotes:
The equation of the asymptotes can be found using the formula y = (b/a)x and y = -(b/a)x.
Plugging in the values, we get the asymptotes: y = (3*sqrt(2/3))/sqrt(2/3) * x and y = -(3*sqrt(2/3))/sqrt(2/3) * x.
Simplifying, the equation of the asymptotes is y = sqrt(2/3) * x and y = -sqrt(2/3) * x.