find the equation of the family of hyperbolas where foci are (~4, 6) & (4, 0). Find the member of the family that passes through (2, 0)
I will assume that you meant:
A(-4,6) and B(4,0) for the foci
Let P(x,y) be any point on the hyperbola
By the basic definition of a hyperbola:
|PA - PB| = 2a, the distance between the two vertices
√( (x+4)^2 + (y-6)^2 ) - √( ( x-4)^2 + y^2) = ±2a
but (2,0) lies on it, so
√( 6^2 + (-6)^2) - √( (-4)^2 + 0) = ±2a
√72 - 4 = ±2a
6√2 - 4 = 2a or 2a = 4-6√2
One wing:
√( (x+4)^2 + (y-6)^2 ) = √( ( x-4)^2 + y^2 ) + (6√2 - 4)
the other wing:
√( (x+4)^2 + (y-6)^2 ) = √( ( x-4)^2 + y^2 ) + (4 - 6√2 )
checking sofar with Wolfram:
http://www.wolframalpha.com/input/?i=plot+%E2%88%9A(+(x%2B4)%5E2+%2B+(y-6)%5E2+)+%3D++%E2%88%9A(+(+x-4)%5E2+%2B+y%5E2+)+%2B+(6%E2%88%9A2+-+4)
looks like a possible focus at (4,0)
http://www.wolframalpha.com/input/?i=plot+%E2%88%9A(+(x%2B4)%5E2+%2B+(y-6)%5E2+)+%3D++%E2%88%9A(+(+x-4)%5E2+%2B+y%5E2+)+%2B+(4+-+6%E2%88%9A2+)
possible focus at (-4,6)
square both sides:
x^2 + 8x+16 + y^2-12y+36 = x^2-8x+16 + (12√2-8)√((x-4)^2+y^2) + 72 - 48√2 + 16
16x - 12y + 36 = 4(3√2-2)√( (x-4)^2 + y^2)
phewww!!
4x - 3y + 9 = (3√2 - 2)√((x-4)^2 + y^2)
square both sides again.
16x^2 - 24xy + 72 + 9y^2 - 54y + 81 =
-12√2 x^2 + 22x^2 + 96√2 x - 176x - 12√2 y^2 + 22y^2 - 192√2 + 352
arhhhhggg!!
I think I made an error somewhere, there has to be an easier way.
Since this is a rotated hyperbola, have you studied the rotation matrix for conics?
Try this:
the centre must be the midpoint of AB which is (0,3)
take the length of AB, divide by 2, that will give you c
from above we found 2a = ....
thus we know "a"
using the property that a^2 + b^2 = c^2, you can find b
so the hyperbola in standard position would be
x^2 / a^2 - (y-3)^2 / b^2 = 1
plug in the values of "a" and "b"
we also know the slope of AB = 6/-8 = -3/4
so tanØ = -3/4
and sinØ = -3/5
and cosØ = 4/5
If I recall, the rotation matrix is
cosØ -sinØ
sinØ cosØ
There has to be examples in your text or your notes.
The last time I did this was about 45 years ago.
To find the equation of the family of hyperbolas, we need to know the standard form of the equation of a hyperbola. The standard form of a hyperbola with the center at (h, k) and the foci at (h + c, k) and (h - c, k) can be given as:
((x - h)² / a²) - ((y - k)² / b²) = 1 (for horizontal hyperbolas)
((y - k)² / a²) - ((x - h)² / b²) = 1 (for vertical hyperbolas)
In this case, the foci are given as (~4, 6) and (4, 0). Since the x-coordinate of the foci is the same for both of them, we can determine that the center of the hyperbola is at (4, 3) (the midpoint of the foci). Additionally, we know that the distance between the foci is equal to 2c, where c is the distance between the center and one of the foci.
To find the value of c, we can use the distance formula:
Distance = √((x2 - x1)² + (y2 - y1)²)
For one of the foci, we can use (4, 0) as (x1, y1), and for the center, we can use (4, 3) as (x2, y2):
c = √((4 - 4)² + (3 - 0)²)
c = √(0 + 9)
c = √9
c = 3
Now that we know c is 3, we can find the equation of the hyperbola. Since the foci are vertically aligned and the center is at (4, 3), the standard form we will use is:
((y - 3)² / a²) - ((x - 4)² / b²) = 1
To find the values of a and b, we need to use the definition of a hyperbola, which states that the sum of the distances from any point on the hyperbola to the two foci is equal to a constant distance (2a in this case).
To find the value of a, we will use the fact that one of the hyperbolas passes through the point (2, 0). Using the distance formula, we can calculate the distance from this point to the foci:
Distance = √((x2 - x1)² + (y2 - y1)²)
For one of the foci, we can use (4, 0) as (x1, y1), and for the point on the hyperbola, we can use (2, 0) as (x2, y2):
2a = √((2 - 4)² + (0 - 0)²)
2a = √((-2)² + 0)
2a = √(4)
2a = 2
Therefore, a = 1.
To find the value of b, we can use the formula:
b = √(c² - a²)
b = √(3² - 1²)
b = √(9 - 1)
b = √8
b = 2√2
Now we have all the values we need to determine the equation of the hyperbola:
((y - 3)² / 1²) - ((x - 4)² / (2√2)²) = 1
Simplifying this equation gives us the equation of the family of hyperbolas:
(y - 3)² - ((x - 4)² / 8) = 1
To find the specific member of the family of hyperbolas that passes through the point (2, 0), we can substitute the values of x and y into the equation and solve for the constant term:
(0 - 3)² - ((2 - 4)² / 8) = 1
(-3)² - (2² / 8) = 1
9 - (4 / 8) = 1
9 - 0.5 = 1
8.5 = 1
Since 8.5 is not equal to 1, the point (2, 0) does not lie on the member of the family of hyperbolas we have found.