Question

# Find the locus of the vertices of the family of parabola $$y=\dfrac { { a }^{ 3 }{ x }^{ 2 } }{ 3 } +\dfrac { { a }^{ 2 }x }{ 2 } -2a$$

Solution

## Family of parabola,$$y=\frac{a^{3}x^{2}}{3} + \frac{a^{2}x}{2} - 2a$$$$h= \frac{-B}{2A} = \frac{-a^{2}/2}{2xa^{3}/3} = \frac{-3}{4a}$$$$k= \frac{(a^{2}/2)^{2} - 4a^{3} (-2a)/3}{4(a^{3}/3)}$$ = $$\frac{a^{4}/4 - a^{4}/3}{4 a^{3}/3}$$$$k= \frac{-35a}{16}$$$$h\times k= \frac{-3}{4a} \times \frac{-35a}{16} = \frac{105}{64} = xy$$Maths

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