JEE Main · 2006 · 30 Apr 2006 · Offline · Parabola

The locus of the vertices of the family of parabolas

y = \frac{a ^{3} x ^{2}}{3} + \frac{a ^{2} x}{2} - 2 a

Answer: (A)

Solution

For a fixed value of

a

the parabola is

y = \frac{a ^{3}}{3} x^{2} + \frac{a ^{2}}{2} x - 2 a

, whose vertex lies at

x = - \frac{a ^{2} /2}{2 ( a ^{3} /3 )} = - \frac{3}{4 a}

. Substituting this back,

y = \frac{a ^{3}}{3} \cdot \frac{9}{16 a ^{2}} + \frac{a ^{2}}{2} \cdot (- \frac{3}{4 a}) - 2 a = \frac{3 a}{16} - \frac{3 a}{8} - 2 a = - \frac{35 a}{16}

. Writing the vertex as

(x, y)

with

x = - \frac{3}{4 a}

and

y = - \frac{35 a}{16}

, the parameter

a

cancels in the product

x y = (- \frac{3}{4 a}) (- \frac{35 a}{16}) = \frac{105}{64}

. The vertices therefore trace the curve

x y = \frac{105}{64}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option