Thread: hyperbola and conic rotation

Alright so im having trouble with 5 problems dealing with hyprebola, and conic rotation If someone can help me that would be great thanks

Question 1
find the canter, vertices, foci and the equations of the asymptotes of the hyperbla and sketch the graph. of x^2-9y^2+36y-72=0

Question 2
Find the standard form of the equation of the hyperbola
Vertices(2,3),(2,-3); Foci (2,6)(2,-6)

Question 3
Find the standard form of the equation of the hyperbola
Vertices(3,0)(3,4); Asymptotes: y=(2/3)x, y=4-(2/3)x

Question 4 (this is dealing with conic rotation i dont know how to do this at all so if someone can explain it a bit that would be nice)
use the discriminant to classify the graph. Then use the quadratic formula to solve for y and use a graphing utility to graph the equation
x^2-8xy+y^2-10x+5y=0

Question 5
Rotate the axes to eliminate the xy-term. Sketch the graph of the resulting equation, showing both sets of axes.
xy-2=0

Hello, ladventchildp!

I'll walk through the first one . . .

1) Find the center, vertices, foci and the equations of the asymptotes
and sketch the graph: .x² - 9y² + 36y - 72 .= .0

Get the equation in standard form . . .

We have: .x² - 9(y² - 4y . . .) .= .72

Then: . . . x² - 9(y² - 4y + 4) .= .72 - 36

. . . . . . . . . . . x² - 9(y - 2)² .= .36

. . . . . . . . . . . x² . (y-2)²
Divide by 36: . --- - ------- .= .1
. . . . . . . . . . .36 . . .4


The hyperbola has center C(0,2) . . . It opens left and right.
. . We have: .a = 6, b = 2

The vertices are: .(±6, 2)

The "focus equation" is: .c² .= .a² + b²
. . So we have: .c² .= .36 + 4 .= .40 . . → . . c .= .±2√10
The foci are: .(±2√10, 2)

The asymptotes are: .y .= .±(1/3}x

Hello, ladventchildp!

If you are expected to do Rotation problems,
. . you should have been given a lot more information and several formulas.

4) Use the discriminant to classify the graph. .Then use the Quadratic Rormula
to solve for y and use a graphing utility to graph the equation.
. . x² - 8xy + y² - 10x + 5y .= .0

The general quadratic equation is: .Ax² + Bxy + Cy² + Dx + Ey + F .= .0

The discriminant is: .d .= .B² - 4AC

. . If d = 0: parabola
. . . .d > 0: hyperbola
. . . .d < 0: ellipse


We have: .A = 1, B = -8, C = 1, D = -10, E = 5, F = 0

The discriminant is: .d .= .(-8)² - 4(1)(1) .= .+60 . . . positive: hyperbola


Write the equation in terms of y:

. . y² - (8x - 5)y + (x² - 10x) .= .0
. . . - . . . . - . . . . . . . . . . . . . . . . . . _____________________
. . . . . . . . . . . . . . . . . . . (8x - 5) ± √(8x - 5)² - 4(1)(x² - 10x)
Quadratic Formula: . y .= . --------------------------------------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2(1)

. . . . . . . . . . . . . . . . . . . . . . . . ______________
. . . . . . . . . . . . . . . (8x - 5) ± √60x² - 40x + 25
And we have: . y .= .--------------------------------
. . . . . . . . . . . . . . . . . . . . . . . .2

5) Rotate the axes to eliminate the xy-term.
Sketch the graph of the resulting equation,
showing both sets of axes: .xy - 2 .= .0

We have: .A = 0, B = 1, C = 0, D = 0, E = 0, F = -2

The discriminant is: .d .= .1² - 4(0)(0) .= .+1 . . . positive: hyperbola

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B
The angle of rotation (θ) is given by: . tan(2θ) .= .-------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C
. . . . . . . . . . . . . . . . . . . . 1
. . We have: . tan(2θ) .= . ------- .= .∞
. . . . . . . . . . . . . . . . . . . 0 - 0

Hence: .2θ .= .π/2 . . → . . θ .= .π/4

. . The hyperbola has been rotated 45° CCW.


How do we "un-rotate" it?

There is another set of formulas:
. . x' .= .x·cosθ - y·sinθ
. . y' .= .x·sinθ + y·cosθ

. . . . . . . . . . . . . . . . . . . . . . . . ._ . - . _ . . - . - . . . _ . . . . _
Since θ = 45°, we have: .x' .= .x/√2 - y/√2, . y' .= .x/√2 + y/√2


Substitute into the original equation: .xy .= .2
. . . . . . . . . . . _ . . . ._ . . . _ . . . . _
We have: . (x/√2 - y/√2)(x/√2 + y/√2) .= .2

. . . . . . . . . . . . . . . . . . . .½x² - ½y² .= .2

Therefore, we have: . . . . . . . . x² - y² .= .4