Question

The locus of the vertices of the family of parabolas \[y=\left[ \dfrac{{{a}^{3}}{{x}^{2}}}{3} \right]+\left( \dfrac{{{a}^{2}}x}{2} \right)-2a\] is
A. \[xy=\dfrac{3}{4}\]
B. \[xy=\dfrac{35}{16}\]
C. \[xy=\dfrac{64}{105}\]
D. \[xy=\dfrac{105}{64}\]

Answer 1

Hint: To solve this type of problem, first we will understand the concept of parabolic equation and then try to simplify the equation given in the question in form of parabolic equation and after solving, substitute the values in required locus equation and you will get your required answer.

Complete step by step answer:
A parabola can be defined as the U shaped plane curve where any point is at an equal distance from a fixed point and that fixed point is known as the focus and the fixed straight line is known as the directrix. We can also say that a parabola is a section of a right circular cone by a plane parallel to the generator of the cone. The simple form of the parabola equation is as: \[{{y}^{2}}=x\] when the directrix is parallel to the \[y\] axis. And the standard equation of parabola when directrix is parallel to \[y\] axis is given as:
\[{{y}^{2}}=4ax\]

Parabolas are used in many fields such as: Mobiles and telecommunication services are possible because of the parabola, in latest dish television facility, In radar dishes and communication satellite, Used in the reflector on torches and spotlights, Also used for focusing the sun’s rays to make a hot spot. Parabolas are competent to concentrate signals to a single point because of the manner they are framed and also because of the characteristics of the focus, that parabolas qualify to get hold of signals and then reflect all of them to a single focus point.

Now, according to the question:Given family of parabolas:
\[y=\left[ \dfrac{{{a}^{3}}{{x}^{2}}}{3} \right]+\left( \dfrac{{{a}^{2}}x}{2} \right)-2a\]
Dividing this whole equation with \[\dfrac{{{a}^{3}}}{3}\]
\[\Rightarrow \dfrac{y}{\dfrac{{{a}^{3}}}{3}}={{x}^{2}}+\dfrac{{{a}^{2}}.3}{2.{{a}^{3}}}x-\dfrac{2a}{\dfrac{{{a}^{3}}}{3}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}={{x}^{2}}+\dfrac{3}{2a}x-\dfrac{6a}{{{a}^{3}}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}={{x}^{2}}+\dfrac{3}{2a}x\]

Now, adding and subtracting \[\dfrac{9}{16{{a}^{2}}}\] on the right hand side.
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}={{x}^{2}}+\dfrac{3}{2a}x+\dfrac{9}{16{{a}^{2}}}-\dfrac{9}{16{{a}^{2}}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}={{x}^{2}}+2\left( \dfrac{3}{2\times 2a}x \right)+\dfrac{9}{16{{a}^{2}}}-\dfrac{9}{16{{a}^{2}}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}={{x}^{2}}+2\left( \dfrac{3}{4a}x \right)+\dfrac{9}{16{{a}^{2}}}-\dfrac{9}{16{{a}^{2}}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}+\dfrac{9}{16{{a}^{2}}}={{x}^{2}}+2\left( \dfrac{3}{4a}x \right)+\dfrac{9}{16{{a}^{2}}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6a}{{{a}^{3}}}+\dfrac{9}{16{{a}^{2}}}={{x}^{2}}+2\left( \dfrac{3}{4a}x \right)+\dfrac{9}{16{{a}^{2}}}\]

Now making perfect square on the right hand side, we will get as:
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{6}{{{a}^{2}}}+\dfrac{9}{16{{a}^{2}}}={{\left( x+\dfrac{3}{4a} \right)}^{2}}\]
\[\Rightarrow \dfrac{3y}{{{a}^{3}}}+\dfrac{105}{16{{a}^{2}}}={{\left( x+\dfrac{3}{4a} \right)}^{2}}\]
\[\Rightarrow {{\left( x+\dfrac{3}{4a} \right)}^{2}}=\dfrac{3y}{{{a}^{3}}}+\dfrac{105}{16{{a}^{2}}}\]
\[\Rightarrow {{\left( x+\dfrac{3}{4a} \right)}^{2}}=\dfrac{3}{{{a}^{3}}}\left( y+\dfrac{35}{16}a \right)\] \[............(1)\]
Let, \[x+\dfrac{3}{4a}=X\] and \[y+\dfrac{35}{16}a=Y\]
So, the equation \[(1)\] reduces to:
\[\Rightarrow {{X}^{2}}=\dfrac{3}{{{a}^{3}}}Y\]

Now, putting \[X=0,Y=0\]
\[\Rightarrow x+\dfrac{3}{4a}=0\]
\[\Rightarrow x=-\dfrac{3}{4a}\]
\[\Rightarrow y+\dfrac{35}{16}a=0\]
\[\Rightarrow y=-\dfrac{35}{16}a\]
Thus, the vertices of parabola are: \[\left( -\dfrac{3}{4a},-\dfrac{35a}{16} \right)\]
Now, required locus: \[xy\]
Substituting values in it and we will get as:
\[\Rightarrow xy=\left( -\dfrac{3}{4a} \right)\times \left( -\dfrac{35}{16}a \right)\]
\[\therefore xy=\dfrac{105}{64}\]
So, the required locus is: \[xy=\dfrac{105}{64}\]

Hence, the correct option is \[D\].

Note: The latus rectum of a parabola can be defined as the chord that passes through the focus and is perpendicular to the axis of the parabola. Two parabolas are said to be equal if their latus rectums are equal. The length of the parabola’s latus rectum is equal to four times the focal length.