Q71 / 150 · JEE Main 2006 · Mathematics
The locus of the vertices of the family of parabolas $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$ is
- Correct answerA$x y=\frac{105}{64}$
- B$x y=\frac{3}{4}$
- C$x y=\frac{35}{16}$
- D$x y=\frac{64}{105}$
Solution
Parabola: $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$
Vertex: $(\alpha, \beta)$
$$
\begin{aligned}
& \alpha=\frac{-a^2 / 2}{2 a^3 / 3}=-\frac{3}{4 a}, \beta=\frac{-\left(\frac{a^4}{4}+4 \cdot \frac{a^3}{3} \cdot 2 a\right)}{4 \frac{a^3}{3}}=-\frac{-\left(\frac{1}{4}+\frac{8}{3}\right) a^4}{\frac{4}{3} a^3} \\
& =-\frac{35}{12} \frac{a}{4} \times 3=-\frac{35}{16} a \\
& \alpha \beta=-\frac{3}{4 a}\left(-\frac{35}{16}\right) a=\frac{105}{64} .
\end{aligned}
$$
Hence, the correct answer is option: 'A'.