Q72 / 155 · WBJEE 2021 · Mathematics

The locus of the vertices of the family of parabolas $6 y=2 a^{3} x^{2}+3 a^{2} x-12 a$ is

  • Correct answer
    A
    $x y=\frac{105}{64}$
  • B
    $x y=\frac{64}{105}$
  • C
    $x y=\frac{35}{16}$
  • D
    $x y=\frac{16}{35}$

Solution

$\begin{aligned} 6 y+12 a &=2 a\left(a^{2} x^{2}+\frac{3}{2} a x\right) \\ &=2 a\left(a^{2} x^{2}+2 \cdot \frac{3}{4} a x+\frac{9}{16}\right)-\frac{9}{8} a \\ \Rightarrow 6 y+\frac{105}{8} a &=2 a\left(a x+\frac{3}{4}\right)^{2} \end{aligned}$ Let Vertices be ( $h, k), \quad h=-\frac{3}{4 a}, k=-\frac{35 a}{16}$ $\therefore \mathrm{hk}=\frac{105}{64}$ Hence, locus is $x y=\frac{105}{64}$

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

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