Q88 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

The combined equation of the diagonals of the parallelogram formed by the lines $\left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32)$ $\left(7 x^2-4 x y+8 y^2\right)=0$ is

  • A
    $x^2-x y+y^2+3 x-15 y=0$
  • B
    $3 x^2-6 x y-2 y^2-15 x-17 y=0$
  • C
    $3 x^2-5 x y-2 y^2-24 x-8 y=0$
  • Correct answer
    D
    None of these

Solution

$\left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32)$ $\left(7 x^2-4 x y+8 y^2\right)=0$ $\Rightarrow\left(7 x^2-4 x y+8 y^2\right)\left(7 x^2-4 x y+8 y^2+4 x-8 y-32\right)=0$ Now, $7 x^2-4 x y+8 y^2+4 x-8 y-32=0$ $\Rightarrow \quad 7 x^2+(-4 y+4) x+8 y^2-8 y-32=0$ Which is quadratic equation in $x$ $\therefore x=\frac{(4 y-4) \pm \sqrt{\left(16 y^2+16-32 y\right)-4 \times 7 \times\left(8 y^2-8 y-32\right)}}{2 \times 7}$ $\Rightarrow x=\frac{(2 y-2) \pm \sqrt{4 y^2+4-32 y-56 y^2+56 y+224}}{7}$ $\Rightarrow x=\frac{(2 y-2) \pm \sqrt{-52 y^2+24 y+228}}{7}$ $=\frac{(2 y-2) \pm \sqrt{-4\left(13 y^2-6 y-57\right)}}{7}$ Here, $D < 0$ So, lines are not possible. Therefore, question is wrong.

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

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