Question

1) Determine the appropriate rotation formulas to use so that the new equation contains no xy- term. (Find the rotation equations for x and y in terms in x' and y. DO NOT FIND THE NEW EQUATION WIHTOUT THE XY-TERMI) 10 pts each a. x2 + 4xy + y2 - 3 = 0 b. 11x2 +103xy + y2 - 4 = 0

Answer 1

The general equation is:

Az? + Bxy + Cy? + Dr + Ey + F = 0........... (1)

For removing the xy term.we have the formulas given below:
Let the new coordinates be x' and y',Thus:
$x=x'cos\theta -y'sin\theta.........(2)$

= 'sino + y cos.... (3

Where $\theta$ is the angle of rotation of the xy coordinate axes.

Thus,to eliminate the term xy from our equation,we have to choose $\theta$ such that:

A-C cot20 .(4) B

Thus,as we have our formulas,we will solve the numericals:

a.

2? + 4xy + y2 – 3 = 0.........(5)

So,comparing the above equation with equation (1),we analyse that:
$A=1,B=4,C=1,D=0,E=0,F=-3$

Thus,Using the equation (4) to find the appropriate $\theta$ ,we get:
$cot2\theta=\frac{1-1}{4}$

$cot2\theta=0$

This will hold if:
$2\theta=\frac{\pi}{2}$

$\mathbf{\theta=\frac{\pi}{4}}$

Thus,we know that:
$cos\frac{\pi}{4}=sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$

Thus,on putting the value in equations (2) and (3),we get:

$x=\frac{x'}{\sqrt{2}}-\frac{y'}{\sqrt{2}}$

$y=\frac{x'}{\sqrt{2}}+\frac{y'}{\sqrt{2}}$

Putting the value of x and y in equation (5),we get:

+ 4(一) +y +y) -3=0 + V2 V2 2 V2

On opening squares and simplifying,we get:
$\frac{1}{2}((x')^2+(y')^2-2x'y'+4(x')^2-4(y')^2+(x')^2+(y')^2+2x'y')-3=0$

$\frac{1}{2}(6(x')^2-2(y')^2)-3=0$

$\mathbf{ANSWER:}$

$\mathbf{3(x')^2-(y')^2-3=0}$

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b:$11x^2+10\sqrt{3}xy+y^2-4=0............(6)$

On comparing the above equation with equation (1),we get:
$A=11,B=10\sqrt{3},C=1,D=0,E=0,F=-4$

Using the formula from equation (4) to find the $\theta$ so that xy term equals to zero,we have:
$cot2\theta=\frac{A-C}{B}$

On putting the values of A C and B,we get:
$cot2\theta=\frac{11-1}{10\sqrt{3}}$

$cot2\theta=\frac{10}{10\sqrt{3}}$

$cot2\theta=\frac{1}{\sqrt{3}}$

Which gets satisfied when:
$2\theta=\frac{\pi}{3}$

$\theta=\frac{\pi}{6}$

Thus,on putting the value of $\theta$ ,we get:
$cos\theta=\frac{\sqrt{3}}{2}$

$sin\theta=\frac{1}{2}$

On putting the values of sin and cos in the equation (2) and (3),we get:

$x=x'cos\theta -y'sin\theta$

$y=x'sin\theta +y'cos\theta$

$x=\frac{\sqrt{3}}{2}x' -\frac{1}{2}y'$

$y=\frac{1}{2}x'+\frac{\sqrt{3}}{2}y'$

On putting the above values of x and y in the equation (6),we get:
$11(\frac{\sqrt{3}}{2}x' -\frac{1}{2}y')^2 +10\sqrt{3}(\frac{\sqrt{3}}{2}x' -\frac{1}{2}y') (\frac{1}{2}x'+\frac{\sqrt{3}}{2}y')+(\frac{1}{2}x' +\frac{\sqrt{3}}{2}y')^2-4=0$

On solving,we get:
$\frac{1}{4}(11(3(x')^2+(y')^2-2\sqrt{3}x'y') +10\sqrt{3}(\sqrt{3}(x')^2+2x'y'-\sqrt{3}(y')^2) +((x')^2+3(y')^2+2\sqrt{3}x'y')-4=0$

$\frac{1}{4}(64(x')^2-16(y')^2)-4=0$

$\mathbf{ANSWER:}$

$\mathbf{16(x')^2-4(y')^2-4=0}$

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