The locus of the vertices of the family of parabolas y = \; a³ x²/3 + \; a² x/2 - 2 a is [AIEEE 2006]

Correct Answer is : xy=10564x y = 105/64xy=64105

Correct Answer is : xy=10564x y = 105/64xy=64105

Test Index

Parabola Part 1

Section: Mathematics

Question : 96 of 100

Marks: +1, -0

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

Solution:

Given parabola is

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

\Rightarrow y = \; \frac{a^3}{3} \left( x^3 + \; \frac{3}{2a} + x + \; \frac{9}{16 a^2} \right) - \; \frac{3a}{16} - 2 a

\Rightarrow y + \frac{35a}{16} = \; \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2

\therefore

Vertex of parabola is

\left( \frac{-3}{4a}, \; \frac{-35a}{16} \right)

To find locus of this vertex,

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

and

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

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

and

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

\Rightarrow \; \frac{-3}{4x} = \; \frac{-16y}{35} \Rightarrow 64 x y = 105

\Rightarrow x y = \frac{105}{64}

which is the required locus.

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