Question

Find the locus of the focus of the family of parabolas \[y=\dfrac{{{a}^{2}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a\].

Answer 1

Hint: Rearrange the terms of the given equation of parabola to write it in the form ${{\left( x-h \right)}^{2}}=4\alpha \left( y-k \right)$. Use the fact that the focus of parabola of the form ${{\left( x-h \right)}^{2}}=4\alpha \left( y-k \right)$ is $\left( h,\alpha +k \right)$. Assume $x=h,y=\alpha +k$ and solve the equations in ‘x’ and ‘y’ to eliminate the variable ‘a’.

Complete step by step answer:
We have the equation of parabola as \[y=\dfrac{{{a}^{2}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a\]. To find the focus of the given family of parabolas, we will rearrange the terms of the given equation of parabola.
Taking LCM of the equation \[y=\dfrac{{{a}^{2}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a\], we have $y=\dfrac{2{{a}^{2}}{{x}^{2}}+3{{a}^{2}}x-12a}{6}$.
Cross multiplying the terms of the above equation, we have $6y=2{{a}^{2}}{{x}^{2}}+3{{a}^{2}}{{x}^{2}}-12a$.
Taking out the common terms, we have $6y=2\left( {{a}^{2}}{{x}^{2}}+\dfrac{3}{2}{{a}^{2}}x-6a \right)$.
Thus, we have $\dfrac{6y}{2}=3y={{a}^{2}}{{x}^{2}}+\dfrac{3}{2}{{a}^{2}}x-6a$
Adding $\dfrac{9{{a}^{2}}}{16}$ on both sides, we have $3y+\dfrac{9{{a}^{2}}}{16}={{a}^{2}}{{x}^{2}}+\dfrac{3}{2}{{a}^{2}}x+\dfrac{9}{16}{{a}^{2}}-6a.....\left( 1 \right)$.
We observe that $\left( {{a}^{2}}{{x}^{2}}+\dfrac{3}{2}{{a}^{2}}x+\dfrac{9{{a}^{2}}}{16} \right)={{\left( ax+\dfrac{3a}{4} \right)}^{2}}$.
So, we can rewrite equation (1) as $3y+\dfrac{9{{a}^{2}}}{16}={{\left( ax+\dfrac{3a}{4} \right)}^{2}}-6a$.
Simplifying the above equation, we have $3y+\dfrac{9{{a}^{2}}}{16}+6a={{\left( ax+\dfrac{3a}{4} \right)}^{2}}$.
Taking out the common terms from right hand side of the above equation, we have $3y+\dfrac{9{{a}^{2}}}{16}+6a={{a}^{2}}{{\left( x+\dfrac{3}{4} \right)}^{2}}$.
Taking out 3 common from left hand side of the above equation, we have $3\left( y+\dfrac{3{{a}^{2}}}{16}+2a \right)={{a}^{2}}{{\left( x+\dfrac{3}{4} \right)}^{2}}$.
Simplifying the above equation, we have $\dfrac{3}{{{a}^{2}}}\left( y+\dfrac{3{{a}^{2}}}{16}+2a \right)={{\left( x+\dfrac{3}{4} \right)}^{2}}$.
We can rewrite the above equation as $\dfrac{3}{{{a}^{2}}}\left( y-\left( -2a-\dfrac{3{{a}^{2}}}{16} \right) \right)={{\left( x+\dfrac{3}{4} \right)}^{2}}$
Multiplying and dividing the left hand side of the given equation by 4, we have $4\left( \dfrac{3}{4{{a}^{2}}} \right)\left( y-\left( -2a-\dfrac{3{{a}^{2}}}{16} \right) \right)={{\left( x+\dfrac{3}{4} \right)}^{2}}$. This is an equation of the form ${{\left( x-h \right)}^{2}}=4\alpha \left( y-k \right)$.
Comparing both the equations, we have $h=\dfrac{-3}{4},k=-2a-\dfrac{3{{a}^{2}}}{16},\alpha =\dfrac{3}{4{{a}^{2}}}$.
We know that the focus of the parabola of the form ${{\left( x-h \right)}^{2}}=4\alpha \left( y-k \right)$ is $\left( h,\alpha +k \right)$.
So, the focus of the parabola $4\left( \dfrac{3}{4{{a}^{2}}} \right)\left( y-\left( -2a-\dfrac{3{{a}^{2}}}{16} \right) \right)={{\left( x+\dfrac{3}{4} \right)}^{2}}$ is $\left( \dfrac{-3}{4},\dfrac{3}{4{{a}^{2}}}+\left( -2a-\dfrac{3{{a}^{2}}}{16} \right) \right)$.
Let’s assume $x=\dfrac{-3}{4},y=\dfrac{3}{4}{{a}^{2}}-2a-\dfrac{3{{a}^{2}}}{16}$.
We will now simplify the equation $x=\dfrac{-3}{4}$.
Multiplying the equation by 4 on both sides, we have $4x=-3$.
Adding 3 on both sides of the above equation, we have $4x+3=3-3=0$.

Hence, the locus of focus of the parabola \[y=\dfrac{{{a}^{2}}{{x}^{2}}}{3}+\dfrac{{{a}^{2}}x}{2}-2a\] is $4x+3=0$.

Note: Focus of a parabola is a fixed point whose distance from each point on the parabola is equal to the distance of the directrix from each point of the parabola. One must know that the focus of the parabola of the form ${{\left( x-h \right)}^{2}}=4\alpha \left( y-k \right)$ is $\left( h,\alpha +k \right)$.