Question

Keeping the origin fixed, the coordinates axes are rotated through an angle $\theta$. If the equation of the line $\frac{x}{a}+\frac{y}{b}=1$ with respect to new axes is $\frac{x}{p}+\frac{y}{q}=1$, then?

• A. $a^2{p}^2+b^2{q}^2=1$
• B. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$
• C. $a^2+b^2=p^2+q^2$
• D. $\frac{1}{a^2}+\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{q^2}$

Answer 1

The correct option is B $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$

$\begin{array}{c}\frac{x}{a}+\frac{y}{b}=1⟶\frac{x}{p}+\frac{x}{q}=1\\ \text{Distance of}pt\cdot (x_1,y_1)\text{from line}xa+by+c=0\text{is}\\ \begin{array}{cc}d=& \frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}\\ \frac{\mid 0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\mid & =\frac{|0+0-1|}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}& \end{array}\\ \text{Hence,}\left(B\right)\text{is the correct option.}\end{array}$