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. $xy=105/64$
• B. $xy=3/4$
• C. $xy=35/16$
• D. $xy=641105$

Answer 1

The correct option is A $xy=105/64$

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

Let the vertex be $(\alpha ,\beta ),$ then

$\alpha =x=\frac{\frac{{-a}^2}{2}}{\frac{2a^3}{3}}=\frac{-3}{4a}$

$\beta =y=\frac{\frac{a^4}{4}+4\ast \frac{a^3}{3}\ast 2a}{4\ast \frac{a^3}{3}}=\frac{(\frac{-1}{4}+\frac{8}{3})a^4}{\frac{4a^3}{3}}$

$\beta =\left(\frac{-35}{16}\right)a$

Now,$\alpha \beta =\left(\frac{-3}{4a}\right)\left(\frac{-35}{16}\right)a$

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

Therefore option (A) is correct.