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}…

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
  1. $\mathrm{an}^2+2 \mathrm{hmn}+\mathrm{bm}^2=0$
  2. $\mathrm{am}^2+2 \mathrm{hmn}+\mathrm{bn}^2=0$
  3. $\mathrm{am}^2-2 \mathrm{hmn}+\mathrm{bn}^2=0$
  4. $\quad \mathrm{an}^2-2 \mathrm{hmn}+\mathrm{bm}^2=0$

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 $\begin{aligned} & a+2 h\left(\frac{n}{m}\right)+b\left(\frac{n}{m}\right)^2=0 \\ & \Rightarrow a^2+2 h m n+n^2=0 \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 2)

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