A line $L$ makes intercepts $a$ and $b$ on the coordinate axes. The axes are rotated through an angle $\theta$ in the positive direction, keeping the origin fixed. If the line $L$ makes intercepts $p$ and $q$ on the new coordinate axes, then $\frac{1}{a^2}+\frac{1}{b^2}=$

• A

  $\frac{1}{p^2 q^2}$
• B

  $\frac{1}{p^2}-\frac{1}{q^2}$
• Correct answer
  C

  $\frac{1}{p^2}+\frac{1}{q^2}$
• D

  $\frac{p q}{p^2+q^2}$

Solution

$$ \text { Let the equation of line is } \frac{x}{a}+\frac{y}{b}=1 \text {. } $$
Distance from origin $$ d=\left|\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right| \Rightarrow \frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2} $$ When line is rotated perpendicular distance from origin is same of line $\frac{x}{p}+\frac{y}{q}=1$ i.e. $d=\left|\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\right| \Rightarrow \frac{1}{d^2}=\frac{1}{p^2}+\frac{1}{q^2}$ From, Eqs. (i) and (ii), we get $$ \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2} $$

Hence, the correct answer is option: 'C'.