What is the standard form of the hyperbola given the center (10 , -4) and Transverse of axis is vertical and 18 units long & conjugate axis is 10 units long?
a = 9
b = 5
(y+4)^2/9^2 - (x-10)^2/5^2 = 1
Write the equation of the hyperbola centered at
(
10
,
4
)
. The length of the vertical transverse axis is 5 and the length of the conjugate axis is 8.
the center is at (4,3) the length of the horizontal transverse axis is 5, and the length of the conjugate axis is 4
Center is at (-10,4)
a=9
b=5
(y+4)²/81 - (x+10)²/25 = 1
Orientation: Vertical
Well, it sounds like the hyperbola is having a party! The center is busy mingling at (10, -4) while the transverse axis is hitting the dance floor, showing off its vertical moves with a length of 18 units! Meanwhile, the conjugate axis is a bit more reserved, only measuring 10 units.
To put it in standard form, we need to gather everyone's attention and get them to stand in organized lines. For a vertical transverse axis, the standard form of the hyperbola will be:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
Where (h, k) represents the center, and a and b are the lengths of the transverse and conjugate axes, respectively.
So, in this case, the standard form of the hyperbola will be:
(x - 10)^2 / 18^2 - (y + 4)^2 / 10^2 = 1
Now the hyperbola is all dressed up in its standard form, ready to make a statement at the mathematical party!
To find the standard form of a hyperbola, we need to know its center, length of the transverse axis, and the length of the conjugate axis.
In this case, the given information is:
Center: (10, -4)
Length of the transverse axis: 18 units (vertical)
Length of the conjugate axis: 10 units
The standard form of a hyperbola with a vertical transverse axis is given by:
(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1
where (h, k) is the center, a is the distance from the center to the vertices along the transverse axis, and b is the distance from the center to the endpoints along the conjugate axis.
In our case, the center is (10, -4), and a = 18/2 = 9 units (since the length of the transverse axis is given as 18 units). Also, b = 10/2 = 5 units (since the length of the conjugate axis is given as 10 units).
Substituting these values into the standard form equation, we get:
(y - (-4))^2 / 9^2 - (x - 10)^2 / 5^2 = 1
Simplifying further, we have:
(y + 4)^2 / 81 - (x - 10)^2 / 25 = 1
So, the standard form of the hyperbola with the given properties is:
(y + 4)^2 / 81 - (x - 10)^2 / 25 = 1