Question

The locus of the point $\left( {a{{\cos }^3}\theta ,b{{\sin }^3}\theta } \right)$ where $0 \leqslant \theta < 2\pi $is
A. ${\left( {{x^2}y} \right)^{\dfrac{2}{3}}} + {\left( {x{y^2}} \right)^{\dfrac{2}{3}}} = 1$
B. ${\left( {{x^2}{y^2}} \right)^{\dfrac{2}{3}}} + {\left( {x{y^2}} \right)^{\dfrac{2}{3}}} = 1$
C. \[{\left( {x/a} \right)^{\dfrac{2}{3}}} + {\left( {y/b} \right)^{\dfrac{2}{3}}} = 1\]
D. \[{\left( {{x^2}/a} \right)^{\dfrac{2}{3}}} + {\left( {{y^2}/b} \right)^{\dfrac{2}{3}}} = 1\]

Answer 1

Hint: Before attempting this question one should have prior knowledge of locus of point and also remember to compare the given coordinates with (x, y) to find the value of x and y and simplify it, use this information to approach towards the solution of the question.

Complete step-by-step answer:
According to the given information we have a point which have coordinates $\left( {a{{\cos }^3}\theta ,b{{\sin }^3}\theta } \right)$ where $0 \leqslant \theta < 2\pi $
Comparing the given coordinates by (x, y) we get
x = $a{\cos ^3}\theta $ and y = $b{\sin ^3}\theta $
Simplifying the value of x
$\dfrac{x}{a}$ = ${\cos ^3}\theta $
$ \Rightarrow $${\left( {\dfrac{x}{a}} \right)^{\dfrac{1}{3}}} = \cos \theta $
Squaring both side in the above equation we get
\[{\left( {{{\left( {\dfrac{x}{a}} \right)}^{\dfrac{1}{3}}}} \right)^2} = {\left( {\cos \theta } \right)^2}\]
$ \Rightarrow $\[{\left( {\dfrac{x}{a}} \right)^{\dfrac{1}{3} \times 2}} = {\cos ^2}\theta \]
$ \Rightarrow $\[{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} = {\cos ^2}\theta \] Taking this equation as equation 1
Now simplifying the value of y
\[\dfrac{y}{b} = {\sin ^3}\theta \]
$ \Rightarrow $\[{\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3}}} = \sin \theta \]
Squaring both side in the above equation
\[{\left( {{{\left( {\dfrac{y}{b}} \right)}^{\dfrac{1}{3}}}} \right)^2} = {\left( {\sin \theta } \right)^2}\]
$ \Rightarrow $\[{\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3} \times 2}} = {\sin ^2}\theta \]
$ \Rightarrow $\[{\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta \] Taking this equation as equation 2
Adding equation 1 and equation 2 we get
\[{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta + {\cos ^2}\theta \]
Since we know that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Therefore \[{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1\]
So the locus of the point is \[{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1\]
Hence option C is the correct option.

Note: In the above solution we used a term “locus” which can be explained as the set of points which satisfies the condition or properties. Let’s explain this concept using an example. Let’s assume that a person walked 3 meter from a point A and now wishes to complete a round around the point A such that he remains 3 meter away from point A until he completes a circular path around point A, so the all points forming the circle are of 3 meter away from point A so these set of all points is called locus of a point.