Q86 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
The slope of one of the pair of lines $2 x^2+h x y+6 y^2=0$ is thrice the slope of the other line then, $h=$
- A$\pm 16$
- B$\pm 9$
- C$\pm 18$
- Correct answerD$\pm 8$
Solution
The equation of given straight line is
$2 x^2+h x y+6 y^2=0 \quad ...\mathrm{(i)}$
Let slope of pair of lines be $m$ and $3 m$ $\therefore$ Pair of equation of straight line $\begin{aligned} & (y-m x)(y-3 m x)=0 \\ & \Rightarrow y^2-4 m x y+3 m^2 x^2=0 \end{aligned}$
On comparing with equation (i), we get $\begin{aligned} & -4 m=\frac{h}{6} \Rightarrow m=\frac{-h}{24} \qquad ...\mathrm{(ii)} \\ & \text { and } 3 m^2=\frac{2}{6} \Rightarrow m^2=\frac{1}{9} \\ & \Rightarrow \frac{h^2}{24 \times 24}=\frac{1}{9} \Rightarrow h^2=\frac{24 \times 24}{9}=64 \\ & \therefore h= \pm 8 . \end{aligned}$
Let slope of pair of lines be $m$ and $3 m$ $\therefore$ Pair of equation of straight line $\begin{aligned} & (y-m x)(y-3 m x)=0 \\ & \Rightarrow y^2-4 m x y+3 m^2 x^2=0 \end{aligned}$
On comparing with equation (i), we get $\begin{aligned} & -4 m=\frac{h}{6} \Rightarrow m=\frac{-h}{24} \qquad ...\mathrm{(ii)} \\ & \text { and } 3 m^2=\frac{2}{6} \Rightarrow m^2=\frac{1}{9} \\ & \Rightarrow \frac{h^2}{24 \times 24}=\frac{1}{9} \Rightarrow h^2=\frac{24 \times 24}{9}=64 \\ & \therefore h= \pm 8 . \end{aligned}$
Hence, the correct answer is option: 'D'.