Question

The locus of the vertices of the family of parabolas y = (a^3x^2)/3 + (a^2x)/2 - 2a is

Answer 1

Answer:

105/64

Step-by-step explanation:

Given The locus of the vertices of the family of parabolas y = (a^3x^2)/3 + (a^2x)/2 - 2a is

We need to find family of parabola, so

y = a^3 x^2 /3 + a^2 x/2 - 2 a

so h = - b / 2 a

  h = - a^2 / 2 / 2 a^3 / 3

h = - a^2 / 2 x 3 / 2 a^3

h = - 3 / 4a

Now k = (a^2 / 2)^2 - 4 a^3(- 2a / 3) / 4(a^3 / 3)

  k = a^4 / 4 - 8 a^4/ 3 / 4a^3 / 3

 3 a^4 + 32 a^4 / 12 / - 4a^3 / 3

k = - 35 a^4 / 12 x 3/4a^3

k = - 35 a / 16

Now h.k = (- 3/4a)(-35 a/16)

(h,k) = 105 / 64