Rotation of Axes Question!

I was given the equation: 2sqrt(2)(x+y)^2=7x+9y

I need to then:

a)Use rotation of axes to show that the following equation represents a parabola

b) Find the XY- and xy- coordinates of the vertex and focus

c) Find the equation of the directrix in XY- and xy-coordinates

Can someone show me how to do this problem so I can do similar problems on my own

Re: Rotation of Axes Question!

ok first let's put our curve in "standard form":

Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0

2√2(x + y)^{2} = 7x + 9y

2√2(x^{2} + 2xy + y^{2}) = 7x + 9y

2√2x^{2} + 4√2xy + 2√2y^{2} - 7x - 9y = 0

so:

A = 2√2

B = 4√2

C = 2√2

D = -7

E = -9

F = 0

now we want to rotate our axes by some angle θ to get a new equation in x' and y':

A'x'^{2} + B'x'y' + C'y'^{2} + D'x' + E'y' + F' = 0

where B' = 0 (eliminating the x'y' term).

after rotating our axes by θ, we can express the new coordinates of a point (x',y') in terms of our old coordinates (x,y) like so:

x' = xcos(θ) + ysin(θ)

y' = ycos(θ) - xsin(θ)

and vice versa:

x = x'cos(θ) - y'sin(θ)

y = x'sin(θ) + y'cos(θ)

let's use the expressions for x and y in terms of x',y', and plug these into our equation:

Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0

so we get:

A(x'cos(θ) - y'sin(θ))^{2} + B(x'cos(θ) - y'sin(θ))(x'sin(θ) + y'cos(θ)) + C(x'sin(θ) + y'cos(θ))^{2} + D(x'cos(θ) - y'sin(θ)) + E(x'sin(θ) + y'cos(θ)) + F = 0

A(x'^{2}cos^{2}(θ) - 2x'y'sin(θ)cos(θ) + y'^{2}sin^{2}(θ))

+ B(x'^{2}sin(θ)cos(θ) + x'y'cos^{2}(θ) - x'y'sin^{2}(θ) - y'^{2}sin(θ)cos(θ))

+ C(x'^{2}sin^{2}(θ) + 2x'y'sin(θ)cos(θ) + y'^{2}cos^{2}(θ))

+ Dx'cos(θ) - Dy'sin(θ) + Ex'sin(θ) + Ey'cos(θ) + F = 0

(whew! that's a lot of algebra!). now let's collect similar terms:

(Acos^{2}(θ) + Bsin(θ)cos(θ) + Csin^{2}(θ))x'^{2}

+ (-2Asin(θ)cos(θ) + Bcos^{2}(θ) - Bsin^{2}(θ) + 2Csin(θ)cos(θ))x'y'

+ (Asin^{2}(θ) - Bsin(θ)cos(θ) + Ccos^{2}(θ))y'^{2}

+ (Dcos(θ) + Esin(θ))x' + (-Dsin(θ) + Ecos(θ))y' + F = 0

this tells us what A',B',C',D',E' and F' are.

remember, we want B' = 0, that is:

-2Asin(θ)cos(θ) + Bcos^{2}(θ) - Bsin^{2}(θ) + 2Csin(θ)cos(θ) = 0

now -2Asin(θ)cos(θ) = -A(2sin(θ)cos(θ)) = -Asin(2θ) and Bcos^{2}(θ) - Bsin^{2}(θ) = B(cos^{2}(θ) - sin^{2}(θ)) = Bcos(2θ)

and 2Csin(θ)cos(θ) = Csin(2θ).

so we have:

(C - A)sin(2θ) + Bcos(2θ) = 0, and re-arranging slightly:

(A - C))sin(2θ) = Bcos(2θ) let's assume for the moment that sin(2θ) ≠ 0. then we can divide by it, to get:

A - C = Bcot(2θ) that is:

cot(2θ) = (A - C)/B <----******important formula #1*******

otherwise, if sin(2θ) = 0, then cos(2θ) ≠ 0, so we have:

(A - C)tan(2θ) = B so

tan(2θ) = B/(A - C) <----******important formula #2*******

since A = C in our equation, we can't use the 2nd formula, so we have:

cot(2θ) = 0, thus cos(2θ) = 0, thus 2θ = π/2, so θ = π/4 (or 45 degrees. note that sin(π/2) = 1 ≠ 0, so we're good).

now we can just plug in the values to find A',C', D', and E' (we already know that F' = F = 0).

from above we have that:

A' = Acos^{2}(θ) + Bsin(θ)cos(θ) + Csin^{2}(θ)

= (2√2)(√2/2)^{2} + (4√2)(√2/2)(√2/2) + (2√2)(√2/2)^{2} = (2√2)(1/2) + (4√2)(1/2) + (2√2)(1/2) = 4√2

and C' = Asin^{2}(θ) - Bsin(θ)cos(θ) + Ccos^{2}(θ)

= (2√2)(√2/2)^{2} -(4√2)(√2/2)(√2/2) + (2√2)(√2/2)^{2} = (2√2)(1/2) - (4√2)(1/2) + (2√2)(1/2) = 0 <---this tells us we have a parabola!!!!

for completeness' sake, we continue:

D' = Dcos(θ) + Esin(θ) = (-7)(√2/2) + (-9)(√2/2) = -8√2

E' = -Dsin(θ) + Ecos(θ) = 7(√2/2) + (-9)(√2/2) = -√2, so our "new equation" is:

4√2x'^{2} - 8√2x' - √2y' = 0. we have a common factor of √2, so let's divide it out:

4x'^{2} - 8x' - y' = 0.

let's put this in a more recognizable form:

y' = 4x'^{2} - 8x'

y' + 4 = 4(x'^{2} - 2x' + 1)

y' = 4(x' - 1)^{2} - 4

from this, we can tell that the vertex is at (1,-4) in x'y'-coordinates. now we use the formulae for expressing x,y in terms of x',y'.

the x-coordinate of the vertex is (1)(√2/2) - (-4)(√2/2) = (5√2)/2

the y-coordinate of the vertex is (1)(√2/2) + (-4)(√2/2) = -(3√2)/2

to find the focus, we write the parabola in another standard form:

(x' - 1)^{2} = 4(1/16)(y' + 4)

this tells us that the focus of the parabola in x'y'-coordinates is at (1,-63/16) and the directrix is the line y' = -65/16.

to find the focus in xy-coordinates, we do the same thing as we did with the vertex:

(1)(√2/2) - (-63/16)(√2/2) = (79√2)/32

(1)(√2/2) + (-63/16)(√2/2) = -(47√2)/32

finding the directrix in xy-coordinates is a bit more troublesome. but let's "cheat" a bit. we know the directrix in x'y'-coordinates has a slope of 0, so in xy-coordinates it has a slope of 1 (tan(π/4) = 1).

so our line is of the form: y - y_{0} = x - x_{0}.

well, we know the point (0,-65/16) is on the line in x'y'-coordinates, so let's find the xy-coordinates of that point, which will give us (x_{0},y_{0}). so we have

(0)(√2/2) - (-65/16)(√2/2) = (65√2)/32 <---x_{0}

(0)(√2/2) + (-65/16)(√2/2) = -(65√2)/32 <---y_{0}.

this tells us the equation of the directrix in xy-coordinates is: y = x - (65√2)/16

(you might well want to check my arithmetic. there has been a lot of calculation, here, and i am only human, and often make mistakes).

Re: Rotation of Axes Question!

The whole point of "rotation of coordinate axes" is to eliminate the "xy" terms- which would appear if you were to multiply out that $\displaystyle (x+ y)^2$. So it should be obvious that x'= (x+ y) will eliminate that immediately! Of course, the line y= -x or x+ y= 0 is perpendicular to y= -x or x- y= 0 so if we let x'= x+y and y'= x- y, we will still have orthogonal axes. Adding those two equations, y'+ x'= 2x so x= (1/2)x'+ (1/2)y'. Subtracting x'- y'= 2y so y= (1/2)x'- (1/2)y'. Now put those into the original equation to get $\displaystyle 2\sqrt{2}x'^2= 7((1/2)x'+ (1/2)y')+ 9((1/2)x'- (1/2)y')= 8x'- y'$.

Of course, y= x and y= -x are at 45 degrees to the x and y axes so that is equivalent to a rotation through 45 degrees.

Re: Rotation of Axes Question!

According to wolfram alpha, the focus is at the point 79/16sqrt(2) and -47/16sqrt(2) and the directrix is at the point x -65/8sqrt(2)

I don't see where the 1/16 comes from while putting it in standard form? And I follow everything up to the point where you compute the focus and directrix of xy and x'y'

Re: Rotation of Axes Question!

1/√2 = √2/2 (as one can verify by cross-multiplying).

ok, i will try to explain where the "1/16" came in.

suppose we have a parabola, just described "geometrically": a parabola is the curve equal distances from a point (the focus) and a line (the directrix). from this description, it is clear the vertex is "halfway between" the focus and the directrix, on the line perpendicular to the directrix through the focus.

for convenience, let us assume the vertex is at (0,0) (if the vertex is at (h,k), then letting x' = x-h, and y' = y-k, we obtain a parabola with vertex at (0,0) in x'y'-coordinates, we can deal with that issue later). we will also assume the the directrix is the line y = -p, so that the focus is the point (0,p).

by definition, the parabola is all points (x,y) such that:

√[(y-p)^{2} + (x-0)^{2}] = √[(y-(-p))^{2} + (x-x)^{2}]

squaring both sides, we have:

y^{2} - 2py + p^{2} + x^{2} = y^{2} + 2py + p^{2}, so:

x^{2} = 4py

so if we have a parabola given in the form: y = ax^{2}, to find "p" (which gives us the focus and directrix) we write

(1/a)y = x^{2}

4(1/(4a))y = x^{2}. this tells us p = 1/(4a).

if instead, our vertex is at (h,k), we have:

4(1/(4a))(y - k) = (x - h)^{2}

now "our a" in the equation above (in your original post) was 4, therefore p = 1/(4*4) = 1/16.

Re: Rotation of Axes Question!

ok yea i got that but wasn't too sure about it when i did it. So, the directrix is at the point y=-p and focus(p,0)

Now, wolfram alpha computes the focus and directrix different from yours. Your work seems to be consistent all throughout, not sure what is wrong.