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Question: The angle of rotation of axes in order to eliminate \(xy\) term in the equation \({{x}^{2}}+2\sqrt{3...

The angle of rotation of axes in order to eliminate xyxy term in the equation x2+23xyy2=2a2{{x}^{2}}+2\sqrt{3}xy-{{y}^{2}}=2{{a}^{2}}is
A.π6\dfrac{\pi }{6}
B.π4\dfrac{\pi }{4}
C.π3\dfrac{\pi }{3}
D.π2\dfrac{\pi }{2}

Explanation

Solution

Here we have to calculate the angle of rotation of axes. For that, we will first the angle of rotation of axes to be ϕ\phi and new coordinates be (x,y)\left( x',y' \right) . We will find the value of old coordinates in terms of new coordinates and ϕ\phi . We will put the value of old coordinates in the given equation and then we will equate the coefficient of xyx'y' with zero. From there, we will get the value of angle of rotation of axes.

Complete step-by-step answer:
Let the angle of rotation of axes be ϕ\phi and let the new coordinates be xx' and yy'.
Thus, the old coordinates are:-
x=xcosϕysinϕx=x'\cos \phi -y'\sin \phi
y=xsinϕ+ycosϕy=x'\sin \phi +y'\cos \phi
Now, we will put the value of coordinates in the given equation x2+23xyy2=2a2{{x}^{2}}+2\sqrt{3}xy-{{y}^{2}}=2{{a}^{2}}
(xcosϕysinϕ)2+23(xcosϕysinϕ)(xsinϕ+ycosϕ)(xsinϕ+ycosϕ)2=2a2{{\left( x'\cos \phi -y'\sin \phi \right)}^{2}}+2\sqrt{3}\left( x'\cos \phi -y'\sin \phi \right)\left( x'\sin \phi +y'\cos \phi \right)-{{\left( x'\sin \phi +y'\cos \phi \right)}^{2}}=2{{a}^{2}}
On further simplification, we get the coefficient of xyx'y' as
23(cos2ϕsin2ϕ)4cosϕsinϕ\Rightarrow 2\sqrt{3}\left( {{\cos }^{2}}\phi -{{\sin }^{2}}\phi \right)-4\cos \phi \sin \phi
Now, we will equate this coefficient with zero to eliminate xyx'y' from the equation.
23(cos2ϕsin2ϕ)4cosϕsinϕ=0........(1)\Rightarrow 2\sqrt{3}\left( {{\cos }^{2}}\phi -{{\sin }^{2}}\phi \right)-4\cos \phi \sin \phi =0........\left( 1 \right)
We know from the trigonometric identities that
cos2ϕsin2ϕ=cos2ϕ\Rightarrow {{\cos }^{2}}\phi -{{\sin }^{2}}\phi =\cos 2\phi and2cosϕsinϕ=sin2ϕ2\cos \phi \sin \phi =\sin 2\phi
We will put these values in equation 1.
23cos2ϕ2sin2ϕ=0\Rightarrow 2\sqrt{3}\cos 2\phi -2\sin 2\phi =0
On further simplifying the terms, we get
tan2ϕ=3\Rightarrow \tan 2\phi =\sqrt{3}
Therefore, the value of ϕ\phi is:-
\Rightarrow 2ϕ=π32\phi =\dfrac{\pi }{3}
\Rightarrow ϕ=π6\phi =\dfrac{\pi }{6}

Hence, the required angle of rotation of axes is π6\dfrac{\pi }{6}.
Thus, the correct option is A.

Note: We had to eliminate the term from the equation x2+23xyy2=2a2{{x}^{2}}+2\sqrt{3}xy-{{y}^{2}}=2{{a}^{2}} to calculate the angle of rotation of axes. So to eliminate any terms from the equation means to equate the coefficient of that term with zero.
Rotation means to turn the figure about a fixed point and that fixed point is called a center of rotation.
The shape and size of the figure remains the same after rotation but its direction gets changed. Rotation can be clockwise or anticlockwise.