hyperbolas?

Question

find the center, vertices, foci, and asymptotes of the hyperbola.
I). [(y^2)/9] - [(x^2)/1] = 1

II). x^2 - 9y^2 + 36y - 72=0

Answer 1

Multiply (8-x)(4-5x)

2. Add (8x^2 -6x+11)+(9x^2+8x-72)

3. The bas of a triangle is 6cm greater than the height. The area is 20cm^2. Find the height and the length of the base.

4. Multiply (y^2-4)(9y^2-6y+9)

5. Express using a positive exponent 1/y-m
1. x^2 - 5x + 6 = 0
2. 2x^2 - 9x - 5 = 0
3. 3x^2 + 5x - 2 = 0
So...
1. g(x)=x^2+12x+k

b^2 - 4ac = 0
(12)^2 - 4(1)(k) = 0
144 - 4k = 0
144 = 4k
36 = k

2. g(x)=kx(x+2)+1
= kx^2 + 2kx + 1

b^2 - 4ac = 0
(2k)^2 - 4(k)(1) = 0
4k^2 - 4k = 0
4k(k - 1) = 0

so, k = 0 or k = 1

3. g(x)=kx(x+4)-1
= kx^2 + 4kx - 1

b^2 - 4ac = 0
(4k)^2 - 4(k)(-1) = 0
16k^2 + 4k = 0
4k(4k + 1) = 0

so, k = 0 or k = -1/4

4. g(x)=3x^2+2kx+6

b^2 - 4ac = 0
(2k)^2 - 4(3)(6) = 0
4k^2 - 72 = 0
4k^2 = 72
k^2 = 18
k = +/- 3 sqrt 2


In each case, the determinat must equal zero

i.e. if ax^2 + bx + c = 0 will have 1 root if and only if

b^2 - 4ac = 0

1
144 - 4(1)(k) = 0 therefore k = 36

2. 4k^2 - 4(k)(1) = 0

Therefore k = 0 and 1

3
16k^2 - 4(k)(-1) = 0

4k^2 +k = 0 therefore k = -0.25 and 0

4.

4k^2 - 4(3)(6) = 0

k = +-sqrt(18) = +-3sqrt(2)

Answer 2

I). y²/9 - x²/1 = 1

The center (h,k) = (0,0).

a² = 9
a = 3

Since the y² variable is positive, the hyperbolas are vertical.
The vertices are:

(h, k - a) and (h, k + a) = (0, -3) and (0, 3)

b² = 1
b = 1

c² = a² + b² = 9 + 1 = 10
c = √10

The foci are:

(h, k - c) and (h, k + c) = (0, -√10) and (0, √10)

The slopes of the asymptotes are
m = ±a/b = ±3/1 = ±3

Since the asymptotes pass thru the center (0,0) their equations are:

y = 3x and y = -3x
______________

II). x² - 9y² + 36y - 72 = 0
x² - 9(y² + 4y) = 72
x² - 9(y² + 4y + 4) = 72 - 9*4 = 72 - 36 = 36
x² - 9(y + 2)² = 36
x²/36 - (y + 2)²/4 = 1

The center of the hyperbolas is (h,k) = (0, -2). The hyperbolas are sideways. You can do the rest.

Answer 3

eh?