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JEE Main 2006
Exam
JEE Main
Year
2006
Subject
Mathematics
Chapter
Parabola

JEE Main 2006 · Mathematics · Parabola

The locus of the vertices of the family of parabolas y=(a^3...

The locus of the vertices of the family of parabolas y=a3x23+a2x2−2ay=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a is

Options

  1. Axy=10564x y=\frac{105}{64}
  2. Bxy=34x y=\frac{3}{4}
  3. Cxy=3516x y=\frac{35}{16}
  4. Dxy=64105x y=\frac{64}{105}

Correct answer

A. xy=10564x y=\frac{105}{64}

Step-by-step solution

Parabola: y=a3x23+a2x2−2ay=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a Vertex: (α,β)(\alpha, \beta)\begin{aligned}& \alpha=\frac{-a^2 / 2}{2 a^3 / 3}=-\frac{3}{4 a}, \beta=\frac{-\left(\frac{a^4}{4}+4 \cdot \frac{a^3}{3} \cdot 2 a\right)}{4 \frac{a^3}{3}}=-\frac{-\left(\frac{1}{4}+\frac{8}{3}\right) a^4}{\frac{4}{3} a^3}\\ & =-\frac{35}{12} \frac{a}{4} \times3=-\frac{35}{16}a \\ & \alpha \beta=-\frac{3}{4 a} \left(-\frac{35}{16}\rright)a=\frac{105}{64} . \end{aligned}$$
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