The coordinate axes are rotated about the origin O in the counter - clock wise direction through an angle ${60}^o$. If 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 $\frac{1}{p^2}+\frac{1}{q^2}=$

$\theta =60°=$Angle of rotation of axes original equation; $x+y-1=0\cdots \cdots \left(1\right)$

Given $p,q$are intercepts made by transformed equation of line(1)

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

$\sin \theta =\sin 60°=\frac{\sqrt{3}}{2};\cos 60°=\cos \theta =\frac{1}{2}$

We know that $x=X\cos \theta -Y\sin \theta \Rightarrow \frac{X-\sqrt{3}Y}{2}$

$y=X\sin \theta +Y\cos \theta \Rightarrow y=\frac{\sqrt{3}X+Y}{2}$

Substituting $x,y$ values in eqn(1)  and simplifying

We get, $X(\frac{1}{2}+\frac{\sqrt{3}}{2})+Y(\frac{-\sqrt{3}}{2}+\frac{1}{2})=1$

$\therefore$ X-intercept $=p=\frac{2}{\sqrt{3}+1};$

Y-intercept $=q=\frac{2}{-\sqrt{3}+1}$

Now $\frac{1}{p^2}+\frac{1}{q^2}=\frac{{(\sqrt{3}+1)}^2}{4}+\frac{{(1-\sqrt{3})}^2}{4}=\frac{2(3+1)}{4}=2$