Q95 / 150 · MHT CET 2021 (24 Sep Shift 2) · Mathematics

If the lines represented by $\left(\mathrm{k}^2+2\right) \mathrm{x}^2+3 \mathrm{xy}-6 \mathrm{y}^2=0$ are perpendicular to each other, then the values of $\mathrm{K}$ are

  • A
    $\pm 3$
  • B
    $\pm 4$
  • C
    $\pm 1$
  • Correct answer
    D
    $\pm 2$

Solution

The lines $\left(k^2+2\right) x^2+3 x y-6 y^2=0$ are perpendicular to each other. $\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2$

Hence, the correct answer is option: 'D'.

Relevant Questions