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).