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# The locus of the vertices of the family ofparabolas$y=\dfrac{{{a}^{3}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a$is(a) $xy=\dfrac{3}{4}$(b) $xy=\dfrac{35}{16}$(c) $xy=\dfrac{64}{105}$(d) $xy=\dfrac{105}{64}$

Last updated date: 20th Mar 2023
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Hint: Vertex of the parabola is the point at which the parabola acquires minimum or maximum value. Differentiate the given equation of parabola and equate it to zero to find the vertex of parabola and then solve it to find the locus of vertices of parabola.

We have the equation of parabola as$y=\dfrac{{{a}^{3}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}- 2a$. To find the vertices of the given family of parabolas, we will differentiate the given equation of
parabola and equate it to zero as vertex of the parabola is the point at which the parabola acquires
minimum or maximum value.
Here, we have$y=\dfrac{{{a}^{3}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a$.
We can write$y$as a sum of three functions$y=f\left( x \right)+g\left( x \right)+h\left( x \right)$such that$f\left( x \right)=\dfrac{{{a}^{3}}{{x}^{2}}}{3},g\left( x \right)=\dfrac{{{a}^{2}}x}{2},h\left( x \right)=-2a$.
We will use sum rule for differentiation of functions which states that if$y=f\left( x \right)+g\left( x \right)+h\left( x \right)$, then we have$\dfrac{dy}{dx}=\dfrac{df\left( x \right)}{dx}+\dfrac{dg\left( x \right)}{dx}+\dfrac{dh\left( x \right)}{dx}$.
Substituting$f\left( x \right)=\dfrac{{{a}^{3}}{{x}^{2}}}{3},g\left( x \right)=\dfrac{{{a}^{2}}x}{2},h\left( x \right)=-2a$in the above equation, we have$\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{{{a}^{3}}{{x}^{2}}}{3} \right)}{dx}+\dfrac{d\left( \dfrac{{{a}^{2}}x}{2} \right)}{dx}+\dfrac{d\left( - 2a \right)}{dx}$. $...\left( 1 \right)$
We know that differentiation of constant function is zero. Thus,$\dfrac{d}{dx}h\left( x \right)=\dfrac{d}{dx}(-2a)=0$ $...\left( 2 \right)$
We know that differentiation of any function of the
form$y=m{{x}^{n}}$is$\dfrac{dy}{dx}=mn{{x}^{n-1}}$.
Substituting$m=\dfrac{{{a}^{3}}}{3},n=2$in the above equation, we have$\dfrac{df\left( x \right)}{dx}=\dfrac{d\left( \dfrac{{{a}^{3}}{{x}^{2}}}{3} \right)}{dx}=\dfrac{2{{a}^{3}}x}{3}$. $...\left( 3 \right)$

Substituting$m=\dfrac{{{a}^{2}}}{2},n=1$in the above equation, we have$\dfrac{dg\left( x \right)}{dx}=\dfrac{d\left( \dfrac{{{a}^{2}}x}{2} \right)}{dx}=\dfrac{{{a}^{2}}}{2}$. $...\left( 4 \right)$
Substituting equation$\left( 2 \right)$,$\left( 3 \right)$and$\left( 4 \right)$in equation$\left( 1 \right)$, we have$\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{{{a}^{3}}{{x}^{2}}}{3} \right)}{dx}+\dfrac{d\left( \dfrac{{{a}^{2}}x}{2} \right)}{dx}+\dfrac{d\left( -2a \right)}{dx}=\dfrac{2{{a}^{3}}x}{3}+\dfrac{{{a}^{2}}}{2}$.
To find the maximum or minimum of a function, we want$\dfrac{dy}{dx}=0$.
Thus, we have$\dfrac{2{{a}^{3}}x}{3}+\dfrac{{{a}^{2}}}{2}=0$
Solving the above equation, we have$x=\dfrac{-3}{4a}$
Substituting the value of$x$in the equation$y=\dfrac{{{a}^{3}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}- 2a$, we have$y=\dfrac{{{a}^{3}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a=\dfrac{{{a}^{3}}{{\left( \dfrac{- 3}{4a} \right)}^{2}}}{3}+\dfrac{{{a}^{2}}\left( \dfrac{-3}{4a} \right)}{2}-2a$.
Solving the above equation, we get$y=\dfrac{{{a}^{3}}}{3}\left( \dfrac{9}{16{{a}^{2}}} \right)+\dfrac{{{a}^{2}}}{2}\left( \dfrac{-3}{4a} \right)-2a=\dfrac{3a}{16}-\dfrac{3a}{8}-2a$.
$\Rightarrow y=\dfrac{3a-6a-32a}{16}=\dfrac{-35a}{16}$
Hence, we have$y=\dfrac{-35a}{16},x=\dfrac{-3}{4a}$.
Multiplying both equations, we get$xy=\dfrac{105}{64}$.
Hence, the correct answer is$xy=\dfrac{105}{64}$.

Note: It’s necessary to consider the fact that the vertex of the parabola is the point at which the parabola acquires minimum or maximum value. We can’t solve this question without using this fact.