Question #b8aa5

1 Answer
Douglas K.
Dec 4, 2017

#x = y^2-2y+1/5#

Explanation:

The reference Rotation of Axes gives equations that will help us to solve this problem.

Given: #x^2-4xy+4y^2+5sqrt(5)y+1=0#

Matching coefficients with the General Cartesian form for a conic section:

#Ax^2+ Bxy + Cy^2+Dx+Ey + F = 0#

We observe that:

#A =1#
#B = -4#
#C = 4#
#D=0#
#E= 5sqrt(5)#
#F = 1#

Equation [9.4.6], in the reference, gives the angle that the parabola must be rotated to align it with one of the axes:

#theta = 1/2tan^-1(B/(C-A))#

#theta = 1/2tan^-1(-4/(4-1))#

#theta = 1/2tan^-1(-4/3)#

My intuition tells me that A' = 0 but we shall use equation (9.4.4a) to verify that:

#A' = (A + C)/2 + [(A - C)/2] cos(2θ) - B/2 sin(2theta)#

#A' = (1 + 4)/2 + [(1 - 4)/2] cos(2(1/2tan^-1(-4/3))) + 4/2 sin(2(1/2tan^-1(-4/3)))#

#A' = 0#

We know that #B' = 0# but feel free to use equation (9.4.4b) to verify that.

To find C', use equation (9.4.4c):

#C' = (A + C)/2 + [(C - A)/2] cos(2θ) + B/2 sin(2θ)#

#C' = (1 + 4)/2 + [(4 - 1)/2] cos(tan^-1(-4/3)) -4/2 sin(tan^-1(-4/3))#

#C' = 5#

To find D', use equation (9.4.4d):

#D' = D cos(θ) - E sin(θ)#

#D' = 5sqrt5sin(1/2tan^-1(-4/3))#

#D' = -5#

To find E', use equation (9.4.4e):

#E' = D sin(θ) + E cos(θ)#

#E' = 5sqrt5cos(1/2tan^-1(-4/3))#

#E' = -10#

Equation (9.4.4f) tells us that #F'=1#

The General Cartesian Form is:

#5y^2 -5x -10y+ 1 = 0#

The standard form for a parabola that opens left or right is:

# x = ay^2+by+c#

The following is the above equation in that form:

#x = y^2-2y+1/5#

Here is an image with the original parabola in red and the unrotated parabola in blue:

www.desmos.com/calculator

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