# Thread: Writing equation for hyperbola

1. ## Writing equation for hyperbola

Write the equation of a hyperbola satisfying the given condition:
with asymptotes y=-x and y=x+2 and vertical tranverse axis length of 4.

vertical tranverse axis

so hyperbola opens up and down

length of 4

so 2b=4, b=2

intersection of y=-x and y=x+2 is (-1,1)

so centre is (-1,1)

The equation so far is:

(x+1)² / a² - (y-1)² /2² = -1

But what is the value of a? How do you find a?

Thanks.

2. Originally Posted by shenton
Write the equation of a hyperbola satisfying the given condition:
with asymptotes y=-x and y=x+2 and vertical tranverse axis length of 4.

vertical tranverse axis

so hyperbola opens up and down

length of 4

so 2b=4, b=2

intersection of y=-x and y=x+2 is (-1,1)

so centre is (-1,1)

The equation so far is:

(x+1)² / a² - (y-1)² /2² = -1

But what is the value of a? How do you find a?

Thanks.
One way:
Solve for y:
y = (1/a)*[a (+/-) 2*sqrt{(x + 1)^2 + a^2}]

For large x this turns into:
y = (1/a)*[a (+/-) 2x] = (+/-)(x/a) + 1 = (+/-)(2x/a)

The asymptotes are (for large x) y = (+/-)x

Thus a = 2.

Below is the graph of (x + 1)^2/4 - (y - 1)^2/4 = -1

-Dan

PS In case anyone saw this whilst I edited it a zillion times, this is my final answer! (And I'm stickin' to it!)

3. Hello, shenton!

Write the equation of a hyperbola satisfying the given condition:
with asymptotes y = -x and y = x + 2 and vertical tranverse axis length of 4.

Vertical tranverse axis, so hyperbola opens up and down.

Length of 4: so 2b = 4, b = 2

Intersection of y = -x and y = x + 2 is (-1,1) . . . so centre is (-1,1) . Good!

The equation so far is:
. . (x + 1)²/a² - (y - 1)²/2² .= .- 1 . Yes!

But what is the value of a?

Recall that the slopes of the asymptotes are: .±b/a

The given asymptotes have slopes ±1
. . Hence: .b .= .±a

Therefore: .a = ±2

The equation is: .(y - 1)²/4 - (x + 1)²/4 .= .1