No need for a full out walktrhough - a brief description of what I need to do will suffice (AKA: enough to get me rolling on the right track, with enough advice to ensure I can do it all)
1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ around the origin have the new coordinates (xcosθ - ysinθ, xsinθ + ycosθ). Draw a picture that clearly illustrates your derivation, and provide plenty of explanation. Hint: consider the distance from the origin to (x,y) as "d", and use the angle sum formulas.
2. Use this result to find the equation of the hyperbola xy = 2 rotated 45° counterclockwise. Compute the vertices, foci, and asymptotes of each of these hyperbolas.
3. Derive the equation of the general conic
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
when rotated counterclockwise through an angle θ. Hint: the coefficient of x^2 becomes (A cos^2θ + Bsinθcosθ + Csin^2θ)
4. Show that rotating the equation through an angle of θ where tan 2θ = B/(A-C) will eliminate that pesky xy term.
5. For each of the following, find:
-the angle of rotation that eliminates the xy term
-the equation resulting from that rotation
-the vertices, foci, and, if appropriate, asymptotes of each equation
a. x^2 + 4xy + y^2 - 3 = 0
b. 3x^2 - 10xy + 3y^2 - 32 = 0
c. x^2 + 4xy + 4y^2 + 5(root5y) + 5 = 0
I know it's lengthy and pretty annoying, but any help at all, no matter how little, is greatly appreciated. Thanks in advance!
Hello, Sconts!
I can help you with #1 . . .
1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ
around the origin have the new coordinates (x·cosθ - y·sinθ, x·sinθ + y·cosθ).
Draw a picture that clearly illustrates your derivation, and provide plenty of explanation.
Hint: consider the distance from the origin to (x,y) as "d", and use the angle-sum formulas.Code:| P' | *(x',y') | * : | * : | * : | d * : P | * : *(x,y) | * d * : | * θ * : : | * * φ : : - * - - - - - - - - + - - + - X O
Point P(x,y) make angle φ with the x-axis.
Point P'(x',y') makes angle P'OP = θ.
Let d = OP = OP'.
The coordinates of P are: .x = d·cosφ, .y = d·sinφ .[1]
The coordinates of P' are: .x' = d·cos(θ + φ), .y' = d·sin(θ + φ)
We have: .x' .= .d·cos(θ + φ) .= .d·cosφ·cosθ - d·sinφ·sinθ
. . . . . . . . . . . . . . . . . . . . . . . . .↓ . . . - . - . . .↓
Substitute [1]: . . . . . . . . .x' .= . .x · cosθ . .- . .y · sinθ
. . Therefore: .x' .= .x·cosθ - y·sinθ
We have: .y' .= .d·sin(θ + φ) .= .d·cosφ·sinθ + d·sinφ·cosθ
. . . . . . . . . . . . . . . . . . . . . . . . .↓ . - . - . - . - .↓
Substitute [1]: . . . . . . . . .y' .= . .x · sinθ . .+ . .y · cosθ
. . Therefore: .y' .= .x·sinθ + y·cosθ