The coordinate axes are rotated about the ori | vedclass

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(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The rotation transformation is given by: $x = x' \cos 60^{\circ} -

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The rotation transformation is given by: $x = x' \cos 60^{\circ} - y' \sin 60^{\circ} = \frac{x'}{2} - \frac{\sqrt{3}y'}{2}$ $y = x' \sin 60^{\circ} + y' \cos 60^{\circ} = \frac{\sqrt{3}x'}{2} + \frac{y'}{2}$ Substituting these into the equation $x+y=1$: $(\frac{x'}{2} - \frac{\sqrt{3}y'}{2}) + (\frac{\sqrt{3}x'}{2} + \frac{y'}{2}) = 1$ $x'(\frac{1+\sqrt{3}}{2}) + y'(\frac{1-\sqrt{3}}{2}) = 1$ This is in the form $\frac{x'}{a} + \frac{y'}{b} = 1$,where $a = \frac{2}{1+\sqrt{3}}$ and $b = \frac{2}{1-\sqrt{3}}$. Then $\frac{1}{a} = \frac{1+\sqrt{3}}{2}$ and $\frac{1}{b} = \frac{1-\sqrt{3}}{2}$. $\frac{1}{a^2} + \frac{1}{b^2} = (\frac{1+\sqrt{3}}{2})^2 + (\frac{1-\sqrt{3}}{2})^2$ $= \frac{1+3+2\sqrt{3}}{4} + \frac{1+3-2\sqrt{3}}{4} = \frac{4+2\sqrt{3} + 4-2\sqrt{3}}{4} = \frac{8}{4} = 2$.

The coordinate axes are rotated about the origin in the counter-clockwise direction through an angle $60^{\circ}$. If $a$ and $b$ are the intercepts made on the new axes by a straight line whose equation referred to the original axes is $x+y=1$,then $\frac{1}{a^2}+\frac{1}{b^2}=$

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