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

The correct option is D

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

Explanation of the correct option.

Compute the locus:

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

Substitute $x$ and $y$ by $h$and $k$ respectively.

$k=\frac{a^3{h}^2}{3}+\frac{a^{2}h}{2}-2a$

$\Rightarrow$ $\frac{3k}{a^3}=h^2+\frac{3h}{2a}-\frac{6}{a^2}$

$\Rightarrow$ $\frac{3k}{a^3}=h^2+\frac{3h}{2a}+\frac{9}{16a^2}-\frac{9}{16a^2}-\frac{6}{a^2}$

$\Rightarrow$$\frac{3}{a^3}\left(k+\frac{35a}{16}\right)={\left(h+\frac{3}{4a}\right)}^2$

The following equation is the equation of the parabola having vertex is $\left(-\frac{3}{4a},-\frac{35a}{16}\right)$

Therefore the locus of vertices is

$kh=\left[\left(-\frac{3}{4a}\right)\times \left(-\frac{35a}{16}\right)\right]$

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

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

Hence, option $\left(D\right)$ is the correct option.