Book Problem

Question

7x^2 - 6sqrt(3)xy + 13y^2 - 16 = 0

 

 

 

Please give me the answer in detail and in graph, been trying to do this for hours and I don't know how to do it 

Answer 1

Jorge, the xy term indicates the rotation of axis. The goal is to choose an angle in which the xy term vanishes in the rotated axis.

 

You start with the formula as cot2θ=(A-C)/B Note: this gives you the angle rotation in which the xy term vanishes.

 

A=7

B=-6√3

C=13

 

This implies cot2θ=(7-13)/-6√3=-6/-6√3=1/√3.

 

2θ=arccot(1/√3) This implies 2θ=60^o. Hence, θ=30^o=pi/6

 

Determine the translated coordinates.

 

x=x'cosθ-y'sinθ=x'cos(pi/6)-y'sin(pi/6)=x'√3/2-y'/2=(x'√3-y')/2

y=x'sinθ+y'cosθ=x'sin(pi/6)+y'cos(pi/6)=(x'+y'√3)/2

 

The easy way to do the problem is to evaluate each term with rotated axis containing x' and y' as new set of coordinates.

 

7x^2=7(x'√3-y')^2/2^2= 21/4*x'^2-7/4*√3x'y'+7/4*y'^2

-6√3xy=-6√3(x'√3-y')(x'+y'√3)/4= -9/2*x'^2-3√3x'y'+3/2*y'^2

13y^2=13(x'+y'√3)^2/2^2= 13/4*x'^2+13/2*√3x'y'+39/4*y'^2

 

Collect the x', x'y', and the y' terms in the original equation.

 

(21/4-9/2+13/4)x'^2 +(-7√3/2-6√3/2+13√3/2)x'y'+(7/4+3/2+39/4)y'^2-16=0

16/4*x'^2+0x'y'+52/4*y'^2-16=0

4x'^2+13y'^2-16=0 

 

4x^2+13y^2=16  Divide both sides of the equation by 16.

 

x'^2/0.25+y'^2/0.8125=1

 

This is an equation of an ellipse on the rotated axis of 30^o and its a vertical ellipse in which its foci points lie on the y' axis. 

 

a^2=0.8125  a≈0.9014

b^2=0.25      b=0.5

 

Hence, the foci is c=±√(0.8125-0.25)=±0.75 in which are (0,0.75) and (0,-0.75)

 

Now, to determine the original coordinates of the foci in which are:

 

x1=0*cos(pi/6)-0.75*sin(pi/6)=-0.75*1/2=-0.375

y1=0*sin(pi/6)+0.75*cos(pi/6)=0.75*√3/2≈0.6495

 

x2=0*cos(pi/6)--0.75sin(pi/6)=0.375

y2=0*sin(pi/6)+-0.75*cos(pi/6)≈-0.6495

 

Therefore, the foci points on the rotated ellipse are (-0.375,0.6495) and (0.375,-0.6495)

 

Apply the formulas to determine where are the major and minor axis of the rotated ellipse is. Here are the steps to sketching the rotated ellipse:

 

1) Draw the original xy-coordinate system.

2) Draw the x'y' coordinate system in which x' axis rotated at 30^o angle with respect to the x axis.

3) Sketch the ellipse on the x'y' coordinate system.

 

Hope this helps. Note: The center of the ellipse is (0,0) in both coordinate systems.

Answer 2

Hey Jorge, are you sure it is 6√3 xy? I'm asking you this because x & y does not multiply with each other:

 

Equation for ellipse [(h,k) will be the center]:

(x-h)²    (y-k)²

-----   +   -----  = 1

  a²         b²

 

Equation for hyperbola [(h,k) will be the center]:

 

(x-h)²    (y-k)²                      (y-h)²      (x-k)²

-----   -   -----  = 1     OR      -------  -   --------  = 1

  a²          b²                            a²           b²

hyperbola that                      hyperbola that

opens vertically                    opens horizontally

 

From these possibilities, there is no way that there will be "xy" multiply together.  Check your "6√3 xy."