What is the locus of the vertices (x, y) of t | vedclass

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(D) The given equation is $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$. To find the vertex,we complete the square for $x$: $y = \frac{a^3}{3} \left( x

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$xy = \frac{105}{64}$

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(D) The given equation is $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$. To find the vertex,we complete the square for $x$: $y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x \right) - 2a$ $y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x + \left(\frac{3}{4a}\right)^2 - \left(\frac{3}{4a}\right)^2 \right) - 2a$ $y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{a^3}{3} \cdot \frac{9}{16a^2} - 2a$ $y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{3a}{16} - 2a$ $y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{35a}{16}$ The vertex $(x, y)$ is given by $x = -\frac{3}{4a}$ and $y = -\frac{35a}{16}$. To find the locus,we multiply $x$ and $y$: $xy = \left( -\frac{3}{4a} \right) \left( -\frac{35a}{16} \right) = \frac{105}{64}$. Thus,the locus is $xy = \frac{105}{64}$.

What is the locus of the vertices $(x, y)$ of the parabolas $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$?

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