Question

If the axes are rotated through an angle $\theta$ and if transformed equation of $ax+by+1=0$ is $px+qy+1=0$, then $a^2-q^2=$

• A. $b^2+p^2$
• B. $p^2-b^2$
• C. $b^2-p^2$
• D. $none$

Answer 1

The correct option is B $p^2-b^2$

$\begin{array}{c}\text{Given,}ax+by+1=0\\ \begin{array}{cc}x& =P\cos \theta +q\sin \theta \\ y& =p\sin \theta +q\mathrm{sos}\theta \end{array}\\ \text{squaring and adding both sides}\\ x^2+y^2=(p\cos \theta -q\sin \theta {)}^2+(p\sin \theta +q\cos \theta {)}^2\\ =p^2{\cos }^2\theta +q^2{\sin }^2\theta +p^2{\sin }^2\theta +q^2{\cos }^2\theta \\ x^2+y^2=p^2+q^2\\ \text{or}{a}^2+b^2=p^2+q^2\\ \Rightarrow a^2-q^2=p^2-b^2\\ \text{hence, (B) is the correct option.}\end{array}$