Question

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

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

Answer 2

Given, $6y=2a^{3}x^{2}+3a^{2}x-12a\quad...\left(1\right)$ 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 $6y=2\left(\frac{-3}{4x}\right)x^{2} + 3\left(-\frac{3}{4x}\right)^{2} -12\left(-\frac{3}{4x}\right)$ $\Rightarrow6y= -\frac{27}{32x}+\frac{27}{16x}+\frac{36}{4x}$ $\Rightarrow 192xy = -27+54+288$ $\Rightarrow xy= \frac{315}{192}$ $= \frac{105}{64}$