Write equation of the hyperbola which satisfies the given conditions:
A.) C(0,0) transverse axis along the x-axis, a focus at (5,0) transverse axis=6.
B.) C(1,-2) transverse axis parallel to the x-axis, transverse axis 6, conjugate axis 10.
C.) C(0,0) transverse axis along the y-axis passing through the points (5,3) and (-3,2).
D.) C(0,6), conjugate axis along the y-axis, asymptotes are 6x-5y+30 and 6x+5y-30=0
A.) To find the equation of the hyperbola when the center is at (0,0), the transverse axis is along the x-axis, and the focus is at (5,0), we can use the standard form equation for a hyperbola:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
Since the center is at (0,0), we have h = 0 and k = 0. The transverse axis length is given as 2a = 6, which means a = 3. The distance from the center to the focus is c = 5.
Using the relationship a^2 + b^2 = c^2, we can solve for b^2:
3^2 + b^2 = 5^2
9 + b^2 = 25
b^2 = 16
Thus, the equation of the hyperbola is:
x^2 / 3^2 - y^2 / 4^2 = 1
B.) The center of the hyperbola is given as C(1,-2), the transverse axis is parallel to the x-axis with a length of 6, and the conjugate axis has a length of 10.
Similar to part A, we can use the standard form equation for a hyperbola:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
With the center at (1,-2), we can now substitute h = 1 and k = -2. The transverse axis length is given as 2a = 6, so a = 3. The conjugate axis length is given as 2b = 10, so b = 5.
Therefore, the equation of the hyperbola is:
(x - 1)^2 / 3^2 - (y + 2)^2 / 5^2 = 1
C.) The center of the hyperbola is given as C(0,0), and the transverse axis is along the y-axis, passing through the points (5,3) and (-3,2).
Again, we'll use the standard form equation for a hyperbola:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
Since the center is at (0,0), we have h = 0 and k = 0. The transverse axis is along the y-axis, so the distance between the two given points is the length of the transverse axis. In this case, it is 3 units.
Therefore, the equation of the hyperbola is:
x^2 / a^2 - y^2 / 3^2 = 1
D.) The center of the hyperbola is given as C(0,6), and the conjugate axis is along the y-axis. The equations of the asymptotes are 6x - 5y + 30 = 0 and 6x + 5y - 30 = 0.
The standard form equation for a hyperbola with the center at (h,k) and the conjugate axis along the y-axis is:
(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1
In this case, the center is at (0,6). So, we can substitute h = 0 and k = 6.
Now, we'll determine the slope of the asymptotes. In general, the slope of the asymptotes for a hyperbola with equations Ax^2 + By^2 + Cx + Dy + E = 0 is given by (B/A)^(1/2). So, in this case, the slopes will be ± (5/6)^(1/2).
Using the slope-intercept form of a line, y = mx + b, we know that the asymptotes have the form:
y = ± (5/6)^(1/2) x + b
Substituting the coordinates of the center (0,6) into these equations, we can solve for b:
6 = ± (5/6)^(1/2) × 0 + b
b = 6
Therefore, the equation of the hyperbola is:
(y - 6)^2 / a^2 - (x - 0)^2 / b^2 = 1
Simplifying further, we have:
(y - 6)^2 / a^2 - x^2 / b^2 = 1
A.) The equation of a hyperbola with center (h, k), transverse axis along the x-axis, a focus at (x1, y1), and transverse axis length 2a is given by:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1
In this case, the center is C(0,0), the focus is (5,0), and the transverse axis length is 6. Therefore, we have:
((x - 0)^2 / (6/2)^2) - ((y - 0)^2 / b^2) = 1
Simplifying, we get:
(x^2 / 3^2) - (y^2 / b^2) = 1
B.) The equation of a hyperbola with center (h, k), transverse axis parallel to the x-axis, transverse axis length 2a, and conjugate axis length 2b is given by:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1
In this case, the center is C(1,-2), the transverse axis length is 6, and the conjugate axis length is 10. Therefore, we have:
((x - 1)^2 / (6/2)^2) - ((y + 2)^2 / (10/2)^2) = 1
Simplifying, we get:
((x - 1)^2 / 3^2) - ((y + 2)^2 / 5^2) = 1
C.) The equation of a hyperbola with center (h, k), transverse axis along the y-axis, and passing through the points (x1, y1) and (x2, y2) is given by:
((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1
In this case, the center is C(0,0), and the hyperbola passes through the points (5,3) and (-3,2). Therefore, we have:
((y - 0)^2 / a^2) - ((x - 0)^2 / b^2) = 1
Using the two given points:
((3 - 0)^2 / a^2) - ((5 - 0)^2 / b^2) = 1
((2 - 0)^2 / a^2) - ((-3 - 0)^2 / b^2) = 1
Simplifying, we get a system of equations:
9 / a^2 - 25 / b^2 = 1
4 / a^2 - 9 / b^2 = 1
Solving this system of equations will give us the values of a and b, which we can then substitute into the general equation.
D.) The equation of a hyperbola with center (h, k), conjugate axis parallel to the y-axis, and asymptotes of the form mx - ny + p = 0 and mx + ny - q = 0 is given by:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1
In this case, the center is C(0,6), and the asymptotes are 6x - 5y + 30 = 0 and 6x + 5y - 30 = 0. Therefore, we have:
((x - 0)^2 / (a/2)^2) - ((y - 6)^2 / (b/2)^2) = 1
Simplifying, we get:
(x^2 / (a/2)^2) - ((y - 6)^2 / (b/2)^2) = 1
You really should review your conic sections, and at least show some effort here.
#A.
c = 5
a = 3 (transverse axis = major axis)
so, b = 4 (conjugate semi-axis)
x^2/9 - y^2/16 = 1
#B. very similar to #A. See what you can do.
#3. You know the equation is
y^2/a^2 - x^2/b^2 = 1
so plug in your points and solve for a,b
#D. the asymptotes have slope b/a. So,
b = 6
a = 5
x^2/25 - (y-6)^2/36 = 1