Question

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

• A. $xy=\frac{3}{4}$
• B. $xy=\frac{35}{16}$
• C. $xy=\frac{64}{105}$
• D. $xy=\frac{105}{64}$

Answer 1

Answer: (D)

The given equation of parabola is
$y=\frac{a^{2}x^{2}}{3}+\frac{a^{2}x}{2}-2a$
$\Rightarrow y+2a=\frac{a^{3}}{3}\left[x^{2}+\frac{3}{2a}x\right]$
$\Rightarrow y+2a=\frac{a^{3}}{3}\left[x^{2}+\frac{3}{2a}x+\frac{9}{16a^{2}}-\frac{9}{16a^{2}}\right]$
$\Rightarrow y+2a=\frac{a^{3}}{3}\left[x+\frac{3}{4a}\right]^{2}-\frac{9}{16a^{2}}\times\frac{a^{3}}{3}$
$\Rightarrow y+2a+\frac{3a}{16}=\frac{a^{3}}{3}\left(x+\frac{3}{4a}\right)^{2}$
$\Rightarrow \left(y+\frac{35a}{16}\right)=\frac{a^{3}}{3}\left(x+\frac{3}{4a}\right)^{2}$
Thus the vertices of parabola is $\left(-\frac{3}{4a}, -\frac{35a}{16}\right)$.
Let $h=-\frac{3}{4a}$
and $k=-\frac{35a}{16}$
$\Rightarrow hk=\frac{105}{64}$
Thus the locus of vertices of a parabola is
$xy=\frac{105}{64}$