Q63 / 150 · MHT CET 2024 (15 May Shift 2) · Mathematics
If one of the lines represented by $a x^2+2 h x y+b y^2=0$ is perpendicular to $\mathrm{m} x+\mathrm{n} y=18$, then
- A$\mathrm{an}^2+2 \mathrm{hmn}+\mathrm{bm}^2=0$
- Correct answerB$\mathrm{am}^2+2 \mathrm{hmn}+\mathrm{bn}^2=0$
- C$\mathrm{am}^2-2 \mathrm{hmn}+\mathrm{bn}^2=0$
- D$\quad \mathrm{an}^2-2 \mathrm{hmn}+\mathrm{bm}^2=0$
Solution
Given equation of pair of lines is $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0$
$\begin{aligned}
& \Rightarrow \mathrm{a}+2 \mathrm{~h}\left(\frac{y}{x}\right)+\mathrm{b}\left(\frac{y}{x}\right)^2=0 \\
& \Rightarrow \mathrm{a}+2 \mathrm{hk}+\mathrm{bk}^2=0 \quad \ldots (i)
\end{aligned}$
Now, slope of line $m x+n y=18$ is $\frac{-m}{n}$ $\therefore \quad$ Slope of the line perpendicular to $\mathrm{m} x+\mathrm{n} y=18$ is $\mathrm{k}=\frac{\mathrm{n}}{\mathrm{m}}$ Substituting the value of $k$ in (i), we get $\begin{aligned} & a+2 h\left(\frac{n}{m}\right)+b\left(\frac{n}{m}\right)^2=0 \\ & \Rightarrow a^2+2 h m n+n^2=0 \end{aligned}$
Now, slope of line $m x+n y=18$ is $\frac{-m}{n}$ $\therefore \quad$ Slope of the line perpendicular to $\mathrm{m} x+\mathrm{n} y=18$ is $\mathrm{k}=\frac{\mathrm{n}}{\mathrm{m}}$ Substituting the value of $k$ in (i), we get $\begin{aligned} & a+2 h\left(\frac{n}{m}\right)+b\left(\frac{n}{m}\right)^2=0 \\ & \Rightarrow a^2+2 h m n+n^2=0 \end{aligned}$
Hence, the correct answer is option: 'B'.