The locus of the vertex of the family of parabolas `y=(a^3x^2)/3+(a^(2x))/2-2a` is `x y=(105)/(64)` (b) `x y=3/4` `x y=(35)/(16)` (d) `x y=(64)/(105)`

Options

A. xy=105/64
B. xy=3/4
C. xy=35/16
D. xy=64/105

Answer

A

Explanation

(1) The family of parabolas is
`y=(a^(3)x^(2))/(3)+(a^(2)x)/(2)-2a=Ax^(2)+Bx+C`
and the vertex is P(-B/2A,-D/4A) `-=` (h,k). Therefore,
`h=-(a^(2)//2)/(2(a^(3)//3))=-(3)/(4a)`
`andk=-((a^(2)//2)^(2)-{4a^(3)(-2a)//3})/(4(a^(3)//3))`
`orh=-(3)/(4a)andk=-(35a)/(16)`
Eliminating a, we have hk =105/64. ltbbrgt Hence, the required is xy=105/64.