Question

when the axes are rotated through an angle π/6 find the transformed equation of x^2+2√3 xy -y^2= 2a^2​​

Answer 1

When X and Y axes are rotated through an angle θ to form X' and Y' axes, then the relation between the axes and the angle is given by,

`\rm{X=X'\cos\theta-Y'\sin\theta}`

`\rm{Y=X'\sin\theta+Y'\cos\theta}`

Then x and y are replaced in terms of x' and y' as,

`\rm{x=x'\cos\theta-y'\sin\theta}`

`\rm{y=x'\sin\theta+y'\cos\theta}`

Here θ = π/6. Then,

`\rm{x=x'\cos\left((\pi)/(6)\right)-y'\sin\left((\pi)/(6)\right)=(x'\sqrt3-y')/(2)}`

`\rm{y=x'\sin\left((\pi)/(6)\right)+y'\cos\left((\pi)/(6)\right)=(x'+y'\sqrt3)/(2)}`

We are asked to find the transformed equation for,

`\longrightarrow\rm{P:x^2+2\sqrt3\,xy-y^2=2a^2}`

We will give substitutions for x and y to find it.

`\small\text{$\longrightarrow\rm{P':\left((x'\sqrt3-y')/(2)\right)^2+2\sqrt3\left((x'\sqrt3-y')/(2)\right)\left((x'+y'\sqrt3)/(2)\right)-\left((x'+y'\sqrt3)/(2)\right)^2=2a^2}$}`

Simplify this equation to get,

`\small\text{$\longrightarrow\rm{P':(x')^2-(y')^2=a^2}$}`

Hence the transformed equation is,

`\longrightarrow\rm{\underline{\underline{x^2-y^2=a^2}}}`