If $X-$intercept is $a$ and $Y-$intercept is $b$ then equation of line on $X-Y$ plane is
$$\frac{x}{a} + \frac{y}{b} = 1 \quad \to (1)$$
We can prove it by considering two points on the line, $(a,0)$ as $(x_1,y_1)$ and $(0,b)$ as $(x_2,y_2)$
So, equation of line will be : $y - y_1 = \left ( \frac{y_2-y_1}{x_2-x_1} \right )(x-x_1)$
$\Rightarrow (y - 0) = \left ( \frac{b-0}{0-a} \right )(x-a)$
$\Rightarrow -ay = bx - ab$
On dividing by $ab$ and rearranging,
$\Rightarrow$ $\frac{x}{a} + \frac{y}{b} = 1$
Now, if a point $(x,y)$ lies on the $X-Y$ plane and if we rotate the co-ordinate system by $\theta$ degree then this point becomes $(x',y')$ on new co-ordinate system and relationship is given by :
$\begin{bmatrix} \cos\theta & \sin\theta \\ - \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} x'\\y' \end{bmatrix}$
So,
- $x' = x\cos\theta +y\sin\theta$ and
- $y'= - x\sin\theta + y\cos\theta$
Now, on the new co-ordinate system, equation on line will be :
$\frac{x'}{a'} + \frac{y'}{b'} = 1$
$\Rightarrow$ $\frac{ x\cos\theta +y\sin\theta}{a'} + \frac{-x\sin\theta + y\cos\theta}{b'} = 1$
$\Rightarrow x\left ( \frac{\cos\theta}{a'} - \frac{\sin\theta}{b'} \right ) + y\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = 1$
On comparing with equation $(1),$
$\left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right ) = \frac{1}{a}$ and
$\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = \frac{1}{b}$
Now, on squaring both sides and adding above $2$ equations :-
$\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = \left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right )^{2} + \left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right )^2$
$ \Rightarrow\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = $ $\frac{\cos^{2}\theta}{a'^2} + \frac{\sin^2\theta}{b'^2} - \frac{2\sin\theta \cos\theta}{a'b'} + \frac{\sin^2\theta}{a'^2} + \frac{\cos^2\theta}{b'^2} + \frac{2\sin\theta \cos\theta}{a'b'}$
$ \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{a'^2} + \frac{1}{b'^2}$
So, Answer is (b)