Question

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

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

Answer 1

The correct option is D $xy=\frac{105}{64}$

$y=\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2a$
$\Rightarrow y=a(\frac{(ax{)}^2}{3}+\frac{ax}{2}-2)$
$\Rightarrow y=a(\frac{(ax{)}^2}{(\sqrt{3}{)}^2}+2\frac{\left(ax\right)}{\left(\sqrt{3}\right)}\cdot \frac{\sqrt{3}}{4}+\frac{3}{4^2}-2-\frac{3}{4^2})$
$\Rightarrow y=a{(\frac{ax}{\sqrt{3}}+\frac{\sqrt{3}}{4})}^2-\frac{35a}{16}$
$\Rightarrow y+\frac{35a}{16}=a{(\frac{ax}{\sqrt{3}}+\frac{\sqrt{3}}{4})}^2\Rightarrow y+\frac{35a}{16}=\frac{a^3}{3}{(x+\frac{3}{4a})}^2$

Vertex is $(\frac{-3}{4a},\frac{-35a}{16})=(h,k)$
Locus is $hk=\frac{3\cdot 35}{4\cdot 16}=\frac{105}{64}$
$xy=\frac{105}{64}$