Question

If by rotation of axes, the expression $a{x^2} + 2hxy + b{y^2}$ be changed to $a'x{'^2} + 2h'x'y' + b'y{'^2}$. Then prove that $a + b = a' + b'$ and $ab - {h^2} = a'b' - h{'^2}$.

Answer 1

Hint: Let \[P\left( {x,y} \right)\] be a point in the \[xy\] plane. If the axes are rotated by an angle \[\theta \] in the anticlockwise direction about the origin, then the coordinates of \[P\] with respect to the rotated axes will be given by the following relations:
\[x = x'\cos \theta - y'\sin \theta \]
\[y = x'\sin \theta + y'\cos \theta \]
Here, \[\left( {x',y'} \right)\]denote the new coordinates of \[P\], here we make use of some trigonometric identities also for further simplification.

Complete answer:
Given the expression $a{x^2} + 2hxy + b{y^2}$ is changed to $a'x{'^2} + 2h'x'y' + b'y{'^2}$ when axes is rotated.
Now using rotation of axes method i.e., when the axes is rotated through an angle $\theta $ then we get,
$x = x'\cos \theta - y'\sin \theta $ and $y = x'\sin \theta + y'\cos \theta $
The new axis of x being inclined at an angle $\theta $ to the old axis, we have to substitute $x = x'\cos \theta - y'\sin \theta $ and $y = x'\sin \theta + y'\cos \theta $ values in the expression $a{x^2} + 2hxy + b{y^2}$ respectively.
Now given expression is $a{x^2} + 2hxy + b{y^2}$
Substituting $x$ and $y$ values in the expression we get,
$ \Rightarrow a{\left( {x'\cos \theta - y'\sin \theta } \right)^2} + 2h\left( {x'\cos \theta - y'\sin \theta } \right)\left( {x'\sin \theta + y'\cos \theta } \right) + b{\left( {x'\sin \theta + y'\cos \theta } \right)^2}$
Now applying square and multiplying we get,
\[a(x{'^2}\cos \theta + y{'^2}\sin \theta - 2x'y'\sin \theta \cos \theta ) + 2h(x{'^2}\sin \theta \cos \theta + x'y'{\cos ^2}\theta - x'y'{\sin ^2}\theta - y{'^2}\sin \theta \cos \theta ) + b(x{'^2}{\sin ^2}\theta + y{'^2}{\cos ^2}\theta + 2x'y'\sin \theta \cos \theta )\]
Now simplifying we get,
\[ax{'^2}\cos \theta + ay{'^2}\sin \theta - 2ax'y'\sin \theta \cos \theta + 2hx{'^2}\sin \theta \cos \theta + hx'y'{\cos ^2}\theta - hx'y'{\sin ^2}\theta - hy{'^2}\sin \theta \cos \theta ) + bx{'^2}{\sin ^2}\theta + by{'^2}{\cos ^2}\theta + 2bx'y'\sin \theta \cos \theta \]
Now transforming this expression the form of $a'x{'^2} + 2h'x'y' + b'y’{^2}$, we get,
\[x{'^2}(a{\cos ^2}\theta + 2h\cos \theta \sin \theta + b{\sin ^2}\theta ) + 2x'y[ - a\cos \theta \sin \theta + h({\cos ^2}\theta - {\sin ^2}\theta ) + b\sin \theta \cos \theta ] + y{'^2}(a{\sin ^2}\theta - 2h\cos \theta \sin \theta + b{\cos ^2}\theta )\]
Now comparing this expression with the expression $a'x{'^2} + 2h'x'y' + b'y{'^2}$, we get,
\[a' = \left( {a{{\cos }^2}\theta + 2h\cos \theta \sin \theta + b{{\sin }^2}\theta } \right) - - - - (1)\]
\[b' = \left( {a{{\sin }^2}\theta - 2h\cos \theta \sin \theta + b{{\cos }^2}\theta } \right) - - - - - (2)\] and,\[h' = \left[ { - a\cos \theta \sin \theta + h\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + b\sin \theta \cos \theta } \right] - - - - (3)\]
Now adding the equations (1) and (2) we get,
\[ \Rightarrow a' + {b'} = a{\cos ^2}\theta + 2h\cos \theta \sin \theta + b{\sin ^2}\theta + a{\sin ^2}\theta - 2h\cos \theta \sin \theta + b{\cos ^2}\theta \]
Now adding the like terms we get,
\[ \Rightarrow a' + {b'} = \left( {a{{\cos }^2}\theta + a{{\sin }^2}\theta } \right) + 2h\cos \theta \sin \theta + \left( {b{{\sin }^2}\theta + b{{\cos }^2}\theta } \right) - 2h\cos \theta \sin \theta \]
Now subtracting the like terms and taking common terms we get,
\[ \Rightarrow a' + {b'} = a\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + b\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\]
Now using the trigonometric identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] and substituting its value in the above equation we get,
\[ \Rightarrow a' + {b'} = a\left( 1 \right) + b\left( 1 \right)\]
Now we get the required result,
\[ \Rightarrow a' + {b'} = a + b\],
Hence proved the first part.
Now for the second part, we have to prove $ab - {h^2} = a'b' - h{'^2}$for this take the equations (1) , (2) and (3) i.e.,
\[b' = \left( {a{{\sin }^2}\theta - 2h\cos \theta \sin \theta + b{{\cos }^2}\theta } \right) - - - - - (2)\] and,\[h' = \left[ { - a\cos \theta \sin \theta + h\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + b\sin \theta \cos \theta } \right] - - - - (3)\]
Now take the equation (1),
\[a' = a{\cos ^2}\theta + 2h\cos \theta \sin \theta + b{\sin ^2}\theta \]
Now multiplying and dividing the equation with 2 we get,
\[a' = \dfrac{1}{2}\left( {2a{{\cos }^2}\theta + 4h\cos \theta \sin \theta + 2b{{\sin }^2}\theta } \right)\]
Now separating two terms we get,
\[a' = \dfrac{1}{2}\left[ {a{{\cos }^2}\theta + a{{\cos }^2}\theta + 4h\cos \theta \sin \theta + b{{\sin }^2}\theta + b{{\sin }^2}\theta } \right]\]
Now using trigonometric identities convert\[\cos \]into\[\sin \]and convert\[\sin \]into\[\cos \]and using \[\sin 2\theta = 2\sin \theta \cos \theta \]we get,
\[a' = \dfrac{1}{2}\left[ {a{{\cos }^2}\theta + a\left( {1 - {{\sin }^2}\theta } \right) + 2h\sin 2\theta + b{{\sin }^2}\theta + b\left( {1 - {{\cos }^2}\theta } \right)} \right]\],
Now expanding the terms in the brackets we get,
\[a' = \dfrac{1}{2}\left[ {a{{\cos }^2}\theta + a - a{{\sin }^2}\theta + 2h\sin 2\theta + b{{\sin }^2}\theta + b - b{{\cos }^2}\theta } \right]\],
Now taking the like terms together we get,
\[a' = \dfrac{1}{2}\left[ {\left( {a + b} \right) + {{\cos }^2}\theta \left( {a - b} \right) - {{\sin }^2}\theta \left( {a - b} \right) + 2h\sin 2\theta } \right]\]
Again taking the like terms we get,
\[a' = \dfrac{1}{2}\left[ {\left( {a + b} \right) + \left( {a - b} \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2h\sin 2\theta } \right]\]
Now using again trigonometric identity we get,
\[a' = \dfrac{1}{2}\left[ {\left( {a + b} \right) + \left( {a - b} \right)\cos 2\theta + 2h\sin 2\theta } \right]\].
Now taking the equation (2) we get,
\[b' = \left( {a{{\sin }^2}\theta - 2h\cos \theta \sin \theta + b{{\cos }^2}\theta } \right) - - - - - (2)\]
Now multiplying and dividing the equation with 2 we get,
\[b' = \dfrac{1}{2}\left( {2a{{\sin }^2}\theta - 4h\cos \theta \sin \theta + 2b{{\cos }^2}\theta } \right)\]
Now separating two terms we get,
\[b' = \dfrac{1}{2}\left[ {a{{\sin }^2}\theta + a{{\sin }^2}\theta - 4h\cos \theta \sin \theta + b{{\cos }^2}\theta + b{{\cos }^2}\theta } \right]\]
Now using trigonometric identities convert\[\cos \]into\[\sin \]and convert\[\sin \]into\[\cos \]and using \[\sin 2\theta = 2\sin \theta \cos \theta \]we get,
\[b' = \dfrac{1}{2}\left[ {a{{\sin }^2}\theta + a\left( {1 - {{\cos }^2}\theta } \right) - 2h\sin 2\theta + b{{\cos }^2}\theta + b\left( {1 - {{\sin }^2}\theta } \right)} \right]\],
Now expanding the terms in the brackets we get,
\[b' = \dfrac{1}{2}\left[ {a{{\sin }^2}\theta + a - a{{\cos }^2}\theta - 2h\sin 2\theta + b{{\cos }^2}\theta + b - b{{\sin }^2}\theta } \right]\],
Now taking the like terms together we get,
\[b' = \dfrac{1}{2}\left[ {\left( {a + b} \right) + {{\sin }^2}\theta \left( {a - b} \right) - {{\cos }^2}\theta \left( {a - b} \right) - 2h\sin 2\theta } \right]\]
Again taking the like terms we get,
\[b' = \dfrac{1}{2}\left[ {\left( {a + b} \right) - \left( {a - b} \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) - 2h\sin 2\theta } \right]\]
Now using again trigonometric identity we get,
\[b' = \dfrac{1}{2}\left[ {\left( {a + b} \right) - \left( {a - b} \right)\cos 2\theta - 2h\sin 2\theta } \right]\].
Now taking the equation (3) i.e.,
\[h' = \left[ { - a\cos \theta \sin \theta + h\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + b\sin \theta \cos \theta } \right] - - - - (3)\]
Now multiplying and dividing the equation with 2 we get,
\[h' = \dfrac{1}{2}\left[ { - 2a\cos \theta \sin \theta + 2h\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2b\sin \theta \cos \theta } \right]\]
Now taking the like terms we get,
\[h' = \dfrac{1}{2}\left[ { - 2\cos \theta \sin \theta \left( {a - b} \right) + 2h\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)} \right]\]
Now using the trigonometric identities we get,
\[h' = \dfrac{1}{2}\left[ { - \sin 2\theta \left( {a - b} \right) + 2h\cos 2\theta } \right]\]
Now we have to prove $ab - {h^2} = a'b' - h’{^2}$ now multiplying and subtracting by substituting the values we got, we get,
\[\left( {\dfrac{1}{2}\left[ {\left( {a + b} \right) + \left( {a - b} \right)\cos 2\theta + 2h\sin 2\theta } \right]} \right)\left( {\dfrac{1}{2}\left[ {\left( {a + b} \right) - \left( {a - b} \right)\cos 2\theta - 2h\sin 2\theta } \right]} \right)- {\left( {\dfrac{1}{2}\left[ { - \sin 2\theta \left( {a - b} \right) + 2h\cos 2\theta } \right]} \right)^2}\]
Now multiplying and simplifying we get,
\[\left( {\dfrac{1}{4}\left[ {{{\left( {a + b} \right)}^2} - {{\left( {\left( {a - b} \right)\cos 2\theta + 2h\sin 2\theta } \right)}^2}} \right]} \right) - \left( {\dfrac{1}{4}\left[ {{{\sin }^2}2\theta {{\left( {a - b} \right)}^2} + 4{h^2}{{\cos }^2}2\theta - 4h\cos 2\theta \left( {a - b} \right)\sin 2\theta } \right]} \right)\]

Now eliminating the terms and grouping the like terms and simplifying we get,
\[\therefore a'b' - h’{^2} = ab - {h^2}\] \[\dfrac{1}{4}\left[ {{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}\left( {{{\sin }^2}2\theta + {{\cos }^2}2\theta } \right) - 4{h^2}\left( {{{\cos }^2}\theta + {{\sin }^2}2\theta } \right)} \right]\]
Now using the trigonometric identity we get,
\[\dfrac{1}{4}\left[ {{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} - 4{h^2}} \right]\],
Now again using the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]and \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] and simplifying we get,
\[\dfrac{1}{4}\left[ {{a^2} + 2ab + {b^2} - {a^2} + 2ab - {b^2} - 4{h^2}} \right]\]
Now eliminating the terms we get,
\[\dfrac{1}{4}\left[ {4ab - 4{h^2}} \right]\]
Now taking 4 common we get,
\[ab - {h^2}\],
\[\therefore a'b' - h{'^2} = ab - {h^2}\],
Hence proved.

If by rotation of axes, the expression $a{x^2} + 2hxy + b{y^2}$ be changed to $a'x’{^2} + 2h'x'y' + b'y’{^2}$, then $a + b = a' + b'$ and $ab - {h^2} = a'b' - h{'^2}$.
Hence proved.

Note:
In these type of questions we use the rotation of axes method i.e., when the axes is rotated through an angle$\theta $then we get,$x = x'\cos \theta - y'\sin \theta $and $y = x'\sin \theta + y'\cos \theta $,the new axis of x being inclined at an angle$\theta $ to the old axis, we have to substitute $x = x'\cos \theta - y'\sin \theta $ and $y = x'\sin \theta + y'\cos \theta $ values., and here using trigonometric identities is also a tricky part as there are many trigonometric identities students must have a clear idea which identity to use where.
Some of the identities and formulas are given below:
${\sin ^2}x + {\cos ^2}x = 1$,
${\tan ^2}x + 1 = {\sec ^2}x$,
$1 + {\cot ^2}x = {\csc ^2}x$
$2\sin x\cos x = \sin 2x$,
${\cos ^2}x - {\sin ^2}x = \cos 2x$
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$.