Question
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) where 0⩽θ<2πis
A. (x2y)32+(xy2)32=1
B. (x2y2)32+(xy2)32=1
C. (x/a)32+(y/b)32=1
D. (x2/a)32+(y2/b)32=1
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) where 0⩽θ<2π
Comparing the given coordinates by (x, y) we get
x = acos3θ and y = bsin3θ
Simplifying the value of x
ax = cos3θ
⇒ (ax)31=cosθ
Squaring both side in the above equation we get
(ax)312=(cosθ)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 $$
$ \Rightarrow {\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3}}} = \sin \theta Squaringbothsideintheaboveequation{\left( {{{\left( {\dfrac{y}{b}} \right)}^{\dfrac{1}{3}}}} \right)^2} = {\left( {\sin \theta } \right)^2}
$ \Rightarrow $$${\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3} \times 2}} = {\sin ^2}\theta
⇒$${\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta Takingthisequationasequation2Addingequation1andequation2weget{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta + {\cos ^2}\theta Sinceweknowthat{\sin ^2}\theta + {\cos ^2}\theta = 1Therefore{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1Sothelocusofthepointis{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1$$
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.