Question

The locus of the vertices of the family of parabolas \(6 y=2 a^{3} x^{2}+3 a^{2} x-12 a\) is
(A) \(x y=\frac{105}{64}\)
(B) \(x y=\frac{64}{105}\)
(C) \(x y=\frac{35}{16}\)
(D) \(x y=\frac{16}{35}\)

Answer 1

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