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Question: The locus of the point \(\left( {a{{\cos }^3}\theta ,b{{\sin }^3}\theta } \right)\) where \(0 \leqsl...

The locus of the point (acos3θ,bsin3θ)\left( {a{{\cos }^3}\theta ,b{{\sin }^3}\theta } \right) where 0θ<2π0 \leqslant \theta < 2\pi is
A. (x2y)23+(xy2)23=1{\left( {{x^2}y} \right)^{\dfrac{2}{3}}} + {\left( {x{y^2}} \right)^{\dfrac{2}{3}}} = 1
B. (x2y2)23+(xy2)23=1{\left( {{x^2}{y^2}} \right)^{\dfrac{2}{3}}} + {\left( {x{y^2}} \right)^{\dfrac{2}{3}}} = 1
C. (x/a)23+(y/b)23=1{\left( {x/a} \right)^{\dfrac{2}{3}}} + {\left( {y/b} \right)^{\dfrac{2}{3}}} = 1
D. (x2/a)23+(y2/b)23=1{\left( {{x^2}/a} \right)^{\dfrac{2}{3}}} + {\left( {{y^2}/b} \right)^{\dfrac{2}{3}}} = 1

Explanation

Solution

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 (acos3θ,bsin3θ)\left( {a{{\cos }^3}\theta ,b{{\sin }^3}\theta } \right) where 0θ<2π0 \leqslant \theta < 2\pi
Comparing the given coordinates by (x, y) we get
x = acos3θa{\cos ^3}\theta and y = bsin3θb{\sin ^3}\theta
Simplifying the value of x
xa\dfrac{x}{a} = cos3θ{\cos ^3}\theta
\Rightarrow (xa)13=cosθ{\left( {\dfrac{x}{a}} \right)^{\dfrac{1}{3}}} = \cos \theta
Squaring both side in the above equation we get
((xa)13)2=(cosθ)2{\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 $$ $ \Rightarrow {\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3}}} = \sin \theta Squaringbothsideintheaboveequation Squaring both side in the above equation {\left( {{{\left( {\dfrac{y}{b}} \right)}^{\dfrac{1}{3}}}} \right)^2} = {\left( {\sin \theta } \right)^2} $ \Rightarrow $$${\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3} \times 2}} = {\sin ^2}\theta
\Rightarrow $${\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta Takingthisequationasequation2Addingequation1andequation2wegetTaking this equation as equation 2 Adding equation 1 and equation 2 we get {\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta + {\cos ^2}\theta Sinceweknowthat Since we know that{\sin ^2}\theta + {\cos ^2}\theta = 1Therefore Therefore{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1Sothelocusofthepointis So the locus of the point is{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1$$
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.