Question

The locus of the vertices of the family of parabola $6y=2a^3{x}^2+3a^{2}x-12a$ is

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

Answer 1

Answer: (A)

Given, $6y=2a^3{x}^2+3a^{2}x-12a...\left(1\right)$
For a vertex of given equation, $\frac{dy}{dx}=0$
$\therefore 6\frac{dy}{dx}=4a^{3}x+3a^2=0$
$\Rightarrow a=-\frac{3}{4x}$
Putting the value of a in $\left(1\right)$, we get
$6y=2\left(\frac{-3}{4x}\right)x^2+3{\left(-\frac{3}{4x}\right)}^2-12\left(-\frac{3}{4x}\right)$
$\Rightarrow 6y=-\frac{27}{32x}+\frac{27}{16x}+\frac{36}{4x}$
$\Rightarrow 192xy=-27+54+288$
$\Rightarrow xy=\frac{315}{192}$
$=\frac{105}{64}$