Question

How do you simplify by rotating the axes at the given equation of 16x^2 -24xy+ 9y^2-250y+1025= 0 ?

`16x^2 -24xy+ 9y^2-250y+1025= 0`

Answer 1

The new equation is `27.52Y^2+200X-150Y+1025=0`

Explanation:

If the equation of the conic is

`Ax^2+Bxy+Cy^2+Dx+Ex+F=0`

The new system of axes is `(X,Y)`

`x=Xcostheta-Ysintheta`

`y=Xsintheta+Ycostheta`

`((x),(y))=((costheta,-sintheta),(sintheta,costheta))*((X),(Y))`

The angle of rotation `theta` is given by

`tan2theta=(B)/(A-C)`

Here, we have

`16x^2-24xy+9y^2-250y+1025=0`.................`(1)`

First calculate,

`B^2-4AC=(-24)^2-4*(16)*(9)=0`

This is the equation of a parabola.

graph{16x^2-24xy+9y^2-250y+1025=0 [-63.8, 102.76, -22.7, 60.7]}

Therefore,

`tan2theta=(-24)/(16-9)=-3.429`

`2theta=106.26^@`

`theta=106.26/2=53.13^@`

Therefore,

`x=Xcos(53.13)-Ysin(53.13)`

`y=Xsin(53.13)+Ycos(53.13)`

`{(x=0.6X-0.8Y),(y=0.8X+0.6Y):}`

Substituting these values in equation `(1)`

`16(0.6X-0.8Y)^2-24(0.6X-0.8Y)(0.8X+0.6Y)+9(0.8X+0.6Y)^2-250(0.8X+0.6Y)+1025=0`

`16*0.36X^2+16*0.64Y^2-16*0.96XY-24(0.48X^2-0.48Y^2-0.28XY)+9(0.64X^2+0.36Y^2+0.96XY)+200X-150Y+1025=0`

`5.76X^2+10.24Y^2-15.36XY-11.52X^2+11.52Y^2+6.72XY+5.76X^2+3.24Y^2+8.64XY+200X-150Y+1025=0`

The new equation is

`27.52Y^2+200X-150Y+1025=0`

graph{27.52y^2-150y+200x+1025=0 [-105.1, 61.46, -24.7, 58.7]}