Question

The line L has intercepts ‘a’ and ‘b’ on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If ‘p’ and ‘q’ are the intercepts of the line L on the new axes, then $\dfrac{1}{{{a}^{2}}}-\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{b}^{2}}}-\dfrac{1}{{{q}^{2}}}$ is equal to
(a) -1
(b) 0
(c) 1
(d) None of these

Answer 1

Hint: First take line as $\dfrac{x}{a}+\dfrac{y}{b}=1$, then if it is rotated by $\alpha $ angle in anticlockwise direction, then replace x by $\left( x\cos \alpha -y\sin \alpha \right)$ and y by $\left( x\sin \alpha +y\cos \alpha \right)$ then put the points (p,0) and (0,q) as they are intercepts to find relation of p and q. Then eliminate $\alpha $ by using identity ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ to get the desired result.

Complete step by step answer:
Let $\alpha $ be the angle by which the line is rotated in anticlockwise direction.
Let the original line be L having the equation,
$\dfrac{x}{a}+\dfrac{y}{b}=1$
So, now rotating by angle $\alpha $ in anticlockwise direction, we will replace x by $\left( x\cos \alpha -y\sin \alpha \right)$ and y by $\left( x\sin \alpha +y\cos \alpha \right)$.
So, the new line L will be,
$\dfrac{\left( x\cos \alpha -y\sin \alpha \right)}{a}+\dfrac{\left( x\sin \alpha +y\cos \alpha \right)}{b}=1$
We are given that p, q are x intercept and y intercepts of new line so it should satisfy this equation, i.e., (p,0) and (0,q) respectively, we get
$\begin{align}
  & \dfrac{\left( p\cos \alpha -0\sin \alpha \right)}{a}+\dfrac{\left( p\sin \alpha +0\cos \alpha \right)}{b}=1 \\
 & \Rightarrow p\left[ \dfrac{\left( \cos \alpha \right)}{a}+\dfrac{\left( \sin \alpha \right)}{b} \right]=1 \\
 & \Rightarrow \left[ \dfrac{\left( \cos \alpha \right)}{a}+\dfrac{\left( \sin \alpha \right)}{b} \right]=\dfrac{1}{p}.......(i) \\
\end{align}$
Similarly,
$\begin{align}
  & \dfrac{\left( 0\cos \alpha -q\sin \alpha \right)}{a}+\dfrac{\left( 0\sin \alpha +q\cos \alpha \right)}{b}=1 \\
 & \Rightarrow q\left[ \dfrac{\left( \cos \alpha \right)}{b}-\dfrac{\left( \sin \alpha \right)}{a} \right]=1 \\
 & \Rightarrow \left[ \dfrac{\left( \cos \alpha \right)}{b}-\dfrac{\left( \sin \alpha \right)}{a} \right]=\dfrac{1}{q}.......(ii) \\
\end{align}$
Now we have to eliminate the terms $\alpha $ . So, we will square the equation (i) and (ii) separately and add them together.
Squaring equation (i), we get
$\Rightarrow {{\left[ \dfrac{\left( \cos \alpha \right)}{a}+\dfrac{\left( \sin \alpha \right)}{b} \right]}^{2}}=\dfrac{1}{{{p}^{2}}}$
Using the formula, ${{\left( c+d \right)}^{2}}={{c}^{2}}+2cd+d$, we get
$\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}{{\cos }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\sin }^{2}}\alpha +\dfrac{2\sin \alpha \cos \alpha }{ab}..........(iii)$
Now squaring equation (ii), we get
$\Rightarrow {{\left[ \dfrac{\left( \cos \alpha \right)}{b}-\dfrac{\left( \sin \alpha \right)}{a} \right]}^{2}}=\dfrac{1}{{{q}^{2}}}$
Using the formula, ${{\left( c+d \right)}^{2}}={{c}^{2}}+2cd+d$, we get
$\dfrac{1}{{{q}^{2}}}=\dfrac{1}{{{a}^{2}}}{{\sin }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\cos }^{2}}\alpha -\dfrac{2\sin \alpha \cos \alpha }{ab}.........(iv)$
Now adding equation (iii) and (iv), we get
$\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}}=\dfrac{1}{{{a}^{2}}}{{\cos }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\sin }^{2}}\alpha +\dfrac{2\sin \alpha \cos \alpha }{ab}+\dfrac{1}{{{a}^{2}}}{{\sin }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\cos }^{2}}\alpha -\dfrac{2\sin \alpha \cos \alpha }{ab}$
Cancelling the like terms, we get
$\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}}=\dfrac{1}{{{a}^{2}}}{{\cos }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\sin }^{2}}\alpha +\dfrac{1}{{{a}^{2}}}{{\sin }^{2}}\alpha +\dfrac{1}{{{b}^{2}}}{{\cos }^{2}}\alpha $
Now grouping and converting we get,
$\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}}=\dfrac{1}{{{a}^{2}}}\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)+\dfrac{1}{{{b}^{2}}}\left( {{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \right)$
Now using identify ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ we get,
$\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{q}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}$
Bringing all the terms on the right hand side, we get
$\dfrac{1}{{{a}^{2}}}-\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{b}^{2}}}-\dfrac{1}{{{q}^{2}}}=0$
So, the value of the given expression $\dfrac{1}{{{a}^{2}}}-\dfrac{1}{{{p}^{2}}}+\dfrac{1}{{{b}^{2}}}-\dfrac{1}{{{q}^{2}}}$ is 0.

Hence the correct answer is option (b).

Note:
Students must be careful while dealing and forming an equation of lines when rotated by any fixed angle. While eliminating also they should be careful about calculation to avoid mistakes.
General mistake that student makes is, after rotating the line they forget to substitute then replace x by $\left( x\cos \alpha -y\sin \alpha \right)$ and y by $\left( x\sin \alpha +y\cos \alpha \right)$
So they won’t obtain the correct answer.