Question

The coordinates axes are rotated about the origin $O$ in the counter clockwise direction through an angle of $\frac{\pi }{6}$. If $a$ and $b$ are intercepts made on the new axes by a straight line whose equation referred to old the axes is $x+y=1$, then the value of $\frac{1}{a^2}+\frac{1}{b^2}$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

Given equation is $x+y=1$
We know that
$x=X\cos \theta -Y\sin \theta$
$y=X\sin \theta +Y\cos \theta$
$\Rightarrow (\cos \theta +\sin \theta )X+(\cos \theta -\sin \theta )Y=1$
$\Rightarrow \frac{(\sqrt{3}+1)}{2}X+\frac{(1-\sqrt{3})}{2}Y=1$ .....(i)
According to problem, we have

$\frac{X}{a}+\frac{Y}{b}=1$ .....(ii)
$\Rightarrow \frac{1}{a}=\frac{(\sqrt{3}+1)}{2}$
$\Rightarrow \frac{1}{b}=\frac{(1-\sqrt{3})}{2}$
So, $\frac{1}{a^2}+\frac{1}{b^2}=2$