Question

Let $0\le \theta \le \dfrac{\pi }{2}$ and $x=X\cos \theta +Y\sin \theta ,y=X\sin \theta -Y\cos \theta $ such that ${{x}^{2}}+4xy+{{y}^{2}}=a{{X}^{2}}+b{{Y}^{2}},$ where a, b are constants then
A. a=-1, b=3
B. $\theta =\dfrac{\pi }{4}$
C. a=3, b=-1
D. $\theta =\dfrac{\pi }{3}$

Answer 1

Hint: We solve this question by substituting the values of x and y given, in the left-hand side of the expression ${{x}^{2}}+4xy+{{y}^{2}}=a{{X}^{2}}+b{{Y}^{2}}.$ Then we simplify the terms on the left-hand side to get only the terms ${{X}^{2}}$ and ${{Y}^{2}}$ in terms of coefficients which should equate to a and b. If we find additional terms, we equate it to 0 and obtain the answer.

Complete step by step answer:
In order to solve this question, let us consider the given values of x and y. We know from the question that,
$\Rightarrow x=X\cos \theta +Y\sin \theta $
$\Rightarrow y=X\sin \theta -Y\cos \theta $
Now, we are also given the expression,
$\Rightarrow {{x}^{2}}+4xy+{{y}^{2}}=a{{X}^{2}}+b{{Y}^{2}}$
We shall solve only the left-hand side first. Substituting the two values for x and y in the above equation on the left-hand side,
$\Rightarrow {{\left( X\cos \theta +Y\sin \theta \right)}^{2}}+4\left( X\cos \theta +Y\sin \theta \right)\left( X\sin \theta -Y\cos \theta \right)+{{\left( X\sin \theta -Y\cos \theta \right)}^{2}}$
Expanding the terms by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab,$ and multiplying the middle terms together,
$\begin{align}
  & \Rightarrow {{X}^{2}}{{\cos }^{2}}\theta +{{Y}^{2}}{{\sin }^{2}}\theta +2XY\sin \theta \cos \theta +4\left( {{X}^{2}}\sin \theta \cos \theta +XY{{\sin }^{2}}\theta -XY{{\cos }^{2}}\theta -{{Y}^{2}}\sin \theta \cos \theta \right)+ \\
 & \text{ }{{X}^{2}}{{\sin }^{2}}\theta +{{Y}^{2}}{{\cos }^{2}}\theta -2XY\sin \theta \cos \theta \\
\end{align}$
Now, we can cancel the third term and the last term as they are the subtraction of the same terms. Then we group ${{X}^{2}},{{Y}^{2}},XY$ terms separately,
$\Rightarrow {{X}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +4\sin \theta \cos \theta \right)+{{Y}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -4\sin \theta \cos \theta \right)+XY\left( 4{{\sin }^{2}}\theta -4{{\cos }^{2}}\theta \right)$
Now, we know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1.$ Using this in the above equation,
$\Rightarrow {{X}^{2}}\left( 1+4\sin \theta \cos \theta \right)+{{Y}^{2}}\left( 1-4\sin \theta \cos \theta \right)+XY\left( 4{{\sin }^{2}}\theta -4{{\cos }^{2}}\theta \right)$
We also know the relation that $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta .$ Using this for the third term,
$\Rightarrow {{X}^{2}}\left( 1+4\sin \theta \cos \theta \right)+{{Y}^{2}}\left( 1-4\sin \theta \cos \theta \right)+XY\left( -4\cos 2\theta \right)$
Now, we shall compare both sides of the equation,
$\Rightarrow {{X}^{2}}\left( 1+4\sin \theta \cos \theta \right)+{{Y}^{2}}\left( 1-4\sin \theta \cos \theta \right)+XY\left( -4\cos 2\theta \right)=a{{X}^{2}}+b{{Y}^{2}}$
As we can see, there are only terms for ${{X}^{2}}$ and ${{Y}^{2}}$ on the right-hand side. Since the right-hand side does not have any $XY$ terms, we equate the $XY$ term on the left-hand side to 0.
$\Rightarrow \left( -4\cos 2\theta \right)=0$
Dividing both sides by -4,
$\Rightarrow \cos 2\theta =0$
We know $\cos x=0$ implies that $x=\dfrac{\pi }{2}$ or any multiple of it. Using the first value of it here,
$\Rightarrow 2\theta =\dfrac{\pi }{2}$
Dividing both sides by 2,
$\Rightarrow \theta =\dfrac{\pi }{4}$
Hence, the value of $\theta =\dfrac{\pi }{4}.$ We now substitute this for the left-hand side of the equation to obtain the values of a and b.
$\Rightarrow {{X}^{2}}\left( 1+4\sin \dfrac{\pi }{4}\cos \dfrac{\pi }{4} \right)+{{Y}^{2}}\left( 1-4\sin \dfrac{\pi }{4}\cos \dfrac{\pi }{4} \right)=a{{X}^{2}}+b{{Y}^{2}}$
We know the value of $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.$ Using this in the above equation,
$\Rightarrow {{X}^{2}}\left( 1+4.\dfrac{1}{\sqrt{2}}.\dfrac{1}{\sqrt{2}} \right)+{{Y}^{2}}\left( 1-4.\dfrac{1}{\sqrt{2}}.\dfrac{1}{\sqrt{2}} \right)=a{{X}^{2}}+b{{Y}^{2}}$
Multiplying the three terms gives us 2,
$\Rightarrow {{X}^{2}}\left( 1+2 \right)+{{Y}^{2}}\left( 1-2 \right)=a{{X}^{2}}+b{{Y}^{2}}$
Adding and subtracting the terms in the brackets,
$\Rightarrow 3{{X}^{2}}-1{{Y}^{2}}=a{{X}^{2}}+b{{Y}^{2}}$
Hence, the values of a=3 and b=-1.

So, the correct answer is “Option B and C”.

Note: We need to know the basic trigonometric formulae and values of standard angles in order to solve such sum. Care must be taken while expanding and simplifying the terms as students tend to miss out a term or two during calculation and it could lead to a wrong answer.