Question

The coordinate axes are rotated through the origin $O$ in the counter clock wise direction through an angle ${60}^0$. lf $p$ and $q$ are the intercepts made on the new axes by a straight line whose equation referred to the original axes is $x+y=1$, then the value of $\frac{1}{p^2}+\frac{1}{q^2}$, is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

Given equation is $x+y=1$

We know that,

$x=X\cos \theta -Y\sin \theta$

$y=X\sin \theta +Y\cos \theta$

$\Rightarrow (\cos \theta +\sin \theta )X+(\cos \theta -\sin \theta )Y=1$

$\Rightarrow \frac{(\sqrt{3}+1)}{2}X+\frac{(1-\sqrt{3})}{2}Y=1$ .....(1)

According to problem, $\frac{X}{p}+\frac{Y}{q}=1$ .....(2)

$\Rightarrow \frac{1}{p}=\frac{(\sqrt{3}+1)}{2}$

And $\frac{1}{q}=\frac{(1-\sqrt{3})}{2}$

So, $\frac{1}{p^2}+\frac{1}{q^2}=2$

Hence, option A is the correct answer.