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.
• B.
• C.
• D.

Answer 1

The correct option is C

Given family of parabolas is y= $\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2a\Rightarrow \frac{y}{\frac{a^3}{3}}=x^2+\frac{a^2}{2}.\frac{3}{a^3}x-\frac{2a}{a\frac{3}{3}}\Rightarrow \frac{3y}{a^3}+\frac{6a}{a^3}=x^2+2\left(\frac{3}{4a}\right)x$
$\Rightarrow {[X+\frac{3}{4a}]}^2=\frac{3y}{a^3}+\frac{6}{a^2}+\frac{9}{16a^2}\Rightarrow {[X+\frac{3}{4a}]}^2=\frac{3y}{a^3}+\frac{105}{16a^2}\Rightarrow {[X+\frac{3}{4a}]}^2=\frac{3}{a^3}[y+\frac{35a}{16}]$
$\Rightarrow \text{Vertex}=[\frac{-3}{4a},\frac{-35a}{16}]$
Let $(x_1,y_1)$ be a point in the locus. Then $x_1=\left(\frac{-3}{4a}\right),y_1=\left(\frac{-35a}{16}\right)$
Now $x_1{y}_1=[-\frac{3}{4a}][-\frac{35a}{16}]\Rightarrow x_1{y}_1=\frac{105}{64}$ $\therefore Equationtothelocusisxy=\frac{105}{64}$