A line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given…
- $a^2+b^2=p^2+q^2$
- $a^2+p^2=b^2+q^2$
- $\frac{1}{a^2}+\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{q^2}$
- $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$
Solution
rotate the axis with respect to angle $\alpha$
Put the values of Eq. (ii) in Eq. (i), we get $ \begin{aligned} & \frac{x \cos \alpha-y \sin \alpha}{a}+\frac{x \sin \alpha+y \cos \alpha}{b}=1 \\ & x\left(\frac{\cos \alpha}{a}+\frac{\sin \alpha}{b}\right)+y\left(\frac{\cos \alpha}{b}-\frac{\sin \alpha}{\alpha}\right)=1 \end{aligned} $ rotated axis $L^{\prime}: \frac{x}{p}+\frac{y}{q}=1$ $\therefore \quad \frac{\cos \alpha}{a}+\frac{\sin \alpha}{b}=\frac{1}{p}$ and $\frac{\cos \alpha}{b}-\frac{\sin \alpha}{a}=\frac{1}{q}$ $\Rightarrow \frac{1}{p^2}+\frac{1}{q^2}=\left(\frac{\cos \alpha}{a}+\frac{\sin \alpha}{b}\right)^2+\left(\frac{\cos \alpha}{b}-\frac{\sin \alpha}{a}\right)^2$ $=\frac{\cos ^2 \alpha}{a^2}+\frac{\sin ^2 \alpha}{b^2}+\frac{2 \sin \alpha \cos \alpha}{a b}+\frac{\cos ^2 \alpha}{b^2}+\frac{\sin ^2 \alpha}{a^2}$ $\frac{-2 \sin \alpha \cos \alpha}{a b}$ $=\frac{\cos ^2 \alpha+\sin ^2 \alpha}{a^2}+\frac{\cos ^2 \alpha+\sin ^2 \alpha}{b^2}$ $\Rightarrow \frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2}$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)