Question

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

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

Answer 1

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

Given the family of parabolas is

$y=\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2a$

Comparing with $y=A{x}^2+Bx+C$,

$A=\frac{a^3}{3},B=\frac{a^2}{2},C=-2a$

Vertex is $(\frac{-B}{2A},\frac{-D}{4A})$

Let the locus of vertex of family of parabolas be $(h,k)$

$\Rightarrow h=\frac{-B}{2A}=\frac{-\frac{a^2}{2}}{2\frac{a^3}{3}}$

$\Rightarrow h=-\frac{3}{4a}$

$\Rightarrow a=-\frac{3}{4h}$ ..........(1)

$\Rightarrow k=\frac{-D}{2A}=\frac{-\frac{a^4}{4}-\frac{4a^3}{3}(-2a)}{2\frac{a^3}{3}}$

$\Rightarrow k=-\frac{35a}{16}$

$\Rightarrow a=-\frac{16k}{35}$ ........(2)

From (1) and (2)

$\Rightarrow \frac{16k}{35}=-\frac{3}{4h}$

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

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