# Rotation of Axes showing parabola

1. May 28, 2012

### indio1

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

Formulas provided:

General Equation of a Conic
Ax^2+Bxy+Cy^2+Dx+Ey+F=0

cot2(fi) =A-C/B

x=Xcos(fi)-Ysin(fi)
y=Xsin(fi)+Ycos(fi)

So, I got up to a certain point but now I am unsure as to what I must do.
I solved for fi, which equals 45 degrees

I solved for x and y which is

x= X/sqrt(2) -Y/sqrt(2)
y= X/sqrt(2)+Y/sqrt(2)

I think I should plug it into the original equation to get x^2=4py or y^2=4px?
Then I am just oblivious as to how I should solve for directrix and vertex,focus for XY, and xy

Any help would be appreciated
This section is new to me so I am working on learning the fundamentals of this section and this touches upon all aspects of the section.

2. May 28, 2012

### tiny-tim

welcome to pf!

hi indio1! welcome to pf!

(have a square-root: √ and a phi: φ and try using the X2 button just above the Reply box )
that's right!

you should get something like X2 + Ax + By = 0

3. May 28, 2012

### indio1

So, i then get this

4x'2 - 8x' - y' = 0.

Put in other form:

y' = 4x'2 - 8x'

y' + 4 = 4(x'2 - 2x' + 1)

y' = 4(x' - 1)2 - 4

vertex is at (1,-4) in x'y'-coordinates. 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

standard form:

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

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

(1)(√2/2) - (-63/16)(√2/2) = (79√2)/32
(1)(√2/2) + (-63/16)(√2/2) = -(47√2)/32

line is of the form: y - y0 = x - x0.

(0)(√2/2) - (-65/16)(√2/2) = (65√2)/32 <---x0
(0)(√2/2) + (-65/16)(√2/2) = -(65√2)/32 <---y0.

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

According to wolframalpha, something is wrong here.
the focus is at the point 79/16sqrt(2) and -47/16sqrt(2) and the directrix is at the point x -65/8sqrt(2)

What is wrong here?

4. May 28, 2012

### tiny-tim

looks fine down to here

(i can't remember anything about focus and directrix, so i can't comment on the rest )