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

Answer:

x^2 - y^2 = a^2

Step-by-step explanation:

x' = xcosA - ysinA

y' = xsinA + ycosA

Here, angle is π/3 or 30°

x' = xcos30° - ysin30°

y' = xsin30° - ycos30°

x' = $\mathrm{\frac{\sqrt3}{2}x - \frac{1}{2}y}$

y' = $\mathrm{\frac{1}{2}x + \frac{\sqrt3}{2}y}$

Hence, now,

$\mathrm{x'^2=\bigg(\frac{\sqrt3}{2}x - \frac{1}{2}y\bigg)^2=\frac{3}{4}x^2-\frac{\sqrt3}{2}xy+\frac{1}{4}y^2}$

$\mathrm{y'^2=\bigg(\frac{\sqrt3}{2}y + \frac{1}{2}x\bigg)^2=\frac{3}{4}y^2+\frac{\sqrt3}{2}xy+\frac{1}{4}x^2}$

$\mathrm{2\sqrt3 xy = 2\sqrt3 \bigg(\frac{\sqrt3}{2}x - \frac{1}{2}y\bigg)\bigg(\frac{1}{2}x + \frac{\sqrt3}{2}y \bigg) = 2\sqrt3\bigg(\frac{\sqrt3}{4}x^2 +\dfrac{1}{2}xy-\frac{\sqrt{3}}{4}y^2\bigg)}$

Now,

x'² + 2√3x'y' - y'²

$\mathrm{\frac{3}{4}x^2-\frac{\sqrt3}{2}xy+\frac{1}{4}y^2 + 2\sqrt3\bigg(\frac{\sqrt3}{4}x^2 +\dfrac{1}{2}xy-\frac{\sqrt{3}}{4}y^2\bigg) - \frac{3}{4}y^2-\frac{\sqrt3}{2}xy-\frac{1}{4}x^2}$

$\mathrm{\frac{3}{4}x^2-\frac{\sqrt3}{2}xy+\frac{1}{4}y^2 + \frac{3}{2}x^2 +\sqrt3xy-\frac{3}{2}y^2 - \frac{3}{4}y^2-\frac{\sqrt3}{2}xy-\frac{1}{4}x^2}$

$\mathrm{\frac{3x^2}{2}+\frac{x^2}{2}-\frac{3y^2}{2}-\frac{y^2}{2}} \\\\ \mathrm{2x^2-2y^2}$

Therefore,

2x^2 - 2y^2 = 2a^2

2( x^2 - y^2 ) = a^2

x^2 - y^2 = a^2