Hyperbolas?

Question

Regarding the hyperbola 4x^2 + 16x - y^2 + 18y +43 = 0.

Specify the vertices, foci, center, and equations of asymptotes. Give the equation in standard form. Leave the points as exact answers, express the asymptotes in slope intercept form...


omg...i'm way in over my head here....please someone help me out ...thank you for your time =)

omg....so sorry that y^2 was supposed to be a 9y^2....ahhh i'm soo dumb..

Answer 1

you have to write that equation in standard form first, which I really don't feel like doin'. sorry. Once you do that, it's easy to identify everything.

Answer 2

1) Take the 43 to the other side of the equation ... it is getting in the way: 4x^2 + 16x - y^2 + 18y = -43

2) Pull out the greatest common factors for the x and y variables: 4(x^2 + 4x) - 1(y^2 - 18y) = -43

3) Complete the square twice ... once for the x variable and again for the y variable: 4(x^2 + 4x + 4) - 1(y^2 - 18y + 81) = -43 + 16 - 81 = -108

4) Every hyperbola equation must equal 1, so divide both sides of the equal sign by -108: 4(x^2 + 4x + 4)/-108 - 1(y^2 - 18y + 81)/-108.

5) Simplify and put positive variable first: (y^2 - 18y + 81)/108 - (x^2 + 4x + 4)/27 = 1

6) Factor both numerators ... each is a perfect square: (y-9)^2/108 - (x+2)^2/27 = 1

7) Center is the opposite of the numbers in the parenthesis. So, (-2, 9) is the center.

That will get you started.

Answer 3

I worked out the radii of the successive circles above circle of radius a million as 3, 17, ninety 9 as reported by means of ????? as below. enable the middle of the 1st circle be (0, a) => its eqn. is x^2 + (y - a)^2 = (a - a million)^2 => x^2 + y^2 - 2ay + 2a - a million = 0 fixing with the eqn. of the hyperbola x^2 - y^2 - a million = 0, removing x, => y^2 - ay + a = 0 As this circle touches the two palms of the hyperbola on the comparable value of y, a^2 - 4a = 0 => a = 4 => radius of the 1st circle is a - a million = 3. enable the middle of the 2d circle be (0, b) => its eqn. is x^2 + (y - b)^2 = (b - 7)^2 => x^2 + y^2 - 2by + 14y - forty 9 = 0 fixing with x^2 - y^2 - a million = 0 => y^2 - by means of + 7by - 24 = 0 For roots of this to be the comparable, b^2 - 28b + ninety six = 0 => (b - 4) (b - 24) = 0 => b = 24 [As b won't be in a position to be 4] => radius of the 2d circle is b - 7 = 17. in addition, enable (0, c) be the middle of the 0.33 circle => its eqn. is x^2 + (y - c)^2 = (c - forty-one)^2 => x^2 + y^2 - 2cy + 82c - 1681 = 0 fixing with x^2 - y^2 - a million = 0 => y^2 - cy + 41c - 840 = 0 => c^2 - 164c + 3360 = 0 => (c - 24) (c - one hundred forty) = 0 => c = one hundred forty and radius of the third circle is c - forty-one = one hundred forty - forty-one = ninety 9. hence, successive circles have radii a million, 3, 17, ninety 9 hence, n = a million => R1 = a million n = 2 => R2 = 3 n = 3 => R3 = 17 n = 4 => R4 = ninety 9 we could desire to continually locate n = n => Rn = ? that's the place i'm caught on the 2d. Will shop attempting. Wolfram alpha generates the radii as below a million, 3, 17, ninety 9, 577, 3363, 19601, 114243, 665857, ....[For extra see link as below.] i got here across the relation between those radii 3 x 6 - a million= 17 17 x 6 - 3 = ninety 9 ninety 9 x 6 - 17 = 577 577x 6 - ninety 9 = 3363 hence, R1 = a million R2 = 3 R3 = 6R2 - R1 R4 = 6R3 - R2 R5 = 6R4 - R3 ...... ..... Rn = 6R(n-a million) - R(n-2) ------------------------------- including, R1 + R2 + ... + Rn = 4 + 6(R3 + R4 + ... + R(n-a million)) - [R1 + R2 + ... + R(n-2)] => R1 + R2 + ... + Rn = 4 - R1 - R2 + 5(R3 + R4 + ... + R(n-a million)) - R(n-2) => R1 + R2 + ... + Rn = 5(R3 + R4 + ... + R(n-a million)) - R(n-2) => 4 + R3 + R4 + ... + R(n-a million) + Rn = 5(R3 + R4 + ... + R(n-a million)) - R(n-2) => 4 + Rn = 4 [R3 + R4 + ... + R(n-a million) - R(n-2) Edit: Sorry i could not get the respond. i hit upon that Quadrillerator has finished it and is the 1st to get to the respond asked. Congrats. TU.

Answer 4

the center is the solution of

8x+16=0
-2y+18 =0 so C(-2,9)
Make the translation of axis X=x+2 and Y=y-9 that will give you
the
equation in standard form
4(X-2)^2+16(X+2) -(Y+9)^2 +18(Y+9)+43 =0
4X^2 +16+32 -Y2 -81+172+43=0
4X^2 -Y^2 =-172
Y^2/172-X2/43=0
The focal axis is the new Y axis a^2+b^2= 172+43= so c^2=215.
You can take it from here( check calculations)

Answer 5

go to the website below and scroll down to 'Two dimensions: Hyperbolas'
http://mathforum.org/dr.math/faq/formulas/faq.ag2.html
hope this helps.