Can you help me solve this hyperbola problem?

Question

4y^2-9x^2+8y-54x-81=0

Answer 1

4y²-9x²+8y-54x-81=0

4(y² + 2y + __ ) - 9(x² + 6x + __) = 81 + __ + __

Once factored as above, you complete the square within each set of parentheses by taking half of the middle term and squaring it. You balance the equation by adding the same thing to the right side of the equation. Be sure to include the factor in front of the parentheses:


4(y² + 2y + 1) - 9(x² + 6x + 9) = 81 + 4 + -81

4(y+1)² -9(x+3)² = 4

Divide by 4 to make equation equal to1

(y+1)² / 1 - (x+3)² / (4/9) = 1

So the center of the hyperbola is (-3,-1) and the curves open up and down, which is called a vertical "transverse" axis.

Therefore the vertices are 1 unit up and down from the center (from the denominator of the positive y term). And the foci are "c" units up and down from the center, where c² = a² + b² (the sum of the denominators in standard form).

Vertices = (-3,0), (-3,-2)
Foci = (-3, -1±√13/3)

Answer 2

4y^2 - 9x^2 + 8y - 54x - 81 = 0
4y^2 + 8y - 9x^2 - 54x - 81 = 0
(4y^2 + 8y) - (9x^2 + 54x) - 81 = 0
4(y^2 + 2y) - 9(x^2 + 6x) - 81 = 0
4(y^2 + 2y + 1 - 1) - 9(x^2 + 6x + 9 - 9) - 81 = 0
4((y + 1)^2 - 1) - 9((x + 3)^2 - 9) - 81 = 0
4(y + 1)^2 - 4 - 9(x + 3)^2 + 81 - 81 = 0
4(y + 1)^2 - 9(x + 3)^2 = 4
(y + 1)^2 - (9/4)(x + 3)^2 = 1
(y + 1)^2 - (((x + 3)^2)/(4/9)) = 1

ANS : (y + 1)^2 - ((9(x + 3)^2)/4) = 1 or to put it in the Hyperbola Formula (y + 1)^2 - (((x + 3)^2)/(4/9)) = 1

Answer 3

(4y^2+8y)+(-9x^2-54x)=81
4(y^2+2y)-9(x^2+6x)=81
Complete the square for x and y.
4(y^2+2y+1)-9(x^2+6x+9)=81+4(1)-9(9)
4(y+1)^2-9(x+3)^2=4
Divide everything by 4.
[(y+1)^2]/1-[9(x+3)^2]/4=1
Move the 9 with the x's underneath the 4.
[(y+1)^2]/1-[(x+3)^2]/(4/9)=1

Answer 4

As I said before, what do you want to know? You figure out the equation of the hyperbola. The equation will tell you how much "tilt" there is in the hyperbola (if you didn't have tilt, you wouldn't have all the linear terms). Then you can figure asymptotes and other weird characteristics of the hyperbola.

Answer 5

ok.. they were given the first one... i will get the subsequent... a) excellent the arrow is shot is 26 through the undeniable fact that is shot even as time = 0, so in simple terms placed a nil in everywhere "t" is in that equation, and also you would possibly want to come across h=26 b) the time it reaches max excellent is even as the spinoff of that excellent vector is 0. So... spinoff is... h/dt=-32t+one hundred fifteen. even as h/dt = 0. you are able to remedy for t. t will equivalent regardless of one hundred fifteen/32 is... about 3.a million seconds or so... c) take the three.a million seconds variety and sub it in for t, on your unique excellent equation... that grant you with the max excellent. looks round 496, in simple terms in my head. d) the time the arrow hits is even as h = 0. so interior the unique equation, set h equivalent to 0 and remedy for t. you receives 2 solutions for t. take the more advantageous one, the different one will be damaging.

Answer 6

What is the question?