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
$xy=\frac{105}{64}$