Question

Transform to axes inclined at ${45}^o$ to the original axes the equations
(1) $x^2-y^2=a^2$,
(2) $17x^2-16xy+17y^2=225$, and

(3) $y^4+x^4+6x^2{y}^2=2$.

Answer 1

Let a point with respect to old axis be $(x,y)$

Let the same point with respect to new axis be $(x^{\prime },y^{\prime })$

If the new axis is formed by rotating old axis by an angle $\theta$ then,

$x=x^{\prime }\cos \theta -y^{\prime }\sin \theta$

$y=x^{\prime }\sin \theta +y^{\prime }\cos \theta$

So, here $\theta ={45}^{\circ }$

$\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$ and $\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$

So, $x=\frac{x^{\prime }}{\sqrt{2}}-\frac{y^{\prime }}{\sqrt{2}}$

and, $y=\frac{x^{\prime }}{\sqrt{2}}+\frac{y^{\prime }}{\sqrt{2}}$

(1) Putting the value of $x$ and $y$ in $x^2-y^2=a^2$, we get

$(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}-x^{\prime }{y}^{\prime })-(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}+x^{\prime }{y}^{\prime })=a^2$

or, $2x^{\prime }{y}^{\prime }+a^2=0$

(2) Putting the value of $x$ and $y$ in $17x^2-16xy+17y^2=225$, we get

$17(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}-x^{\prime }{y}^{\prime })-16((\frac{x^{\prime }}{\sqrt{2}}-\frac{y^{\prime }}{\sqrt{2}})(\frac{x^{\prime }}{\sqrt{2}}+\frac{y^{\prime }}{\sqrt{2}})+17(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}+x^{\prime }{y}^{\prime }))=225$

or, $17(x^{\prime 2}+y^{\prime 2})-16(\frac{x^{\prime 2}}{2}-\frac{y^{\prime 2}}{2})=225$

or, $9x^{\prime 2}+25y^{\prime 2}=225$

(3)$y^4+x^4+6x^2{y}^2=2$ can be written as,

$(x^2-y^2{)}^2+8x^2{y}^2=2$

Putting the value of $x$ and $y$ in $(x^2-y^2{)}^2+8x^2{y}^2=2$,we get

${((\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}-x^{\prime }{y}^{\prime })-(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}+x^{\prime }{y}^{\prime }))}^2+8(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}-x^{\prime }{y}^{\prime })(\frac{x^{\prime 2}}{2}+\frac{y^{\prime 2}}{2}+x^{\prime }{y}^{\prime })=2$

or, $(-2x^{\prime }{y}^{\prime }{)}^2+8(\frac{x^{\prime 4}}{4}+\frac{y^{\prime 4}}{4}-\frac{x^{\prime 2}{y}^{\prime 2}}{2})$=2

or, $4x^{\prime 2}{y}^{\prime 2}+2x^{\prime 4}+2y^{\prime 2}-4x^{\prime 2}{y}^{\prime 2}=2$

or, $x^{\prime 4}+y^{\prime 4}=1$