Question

Find the equation of locus of the moving point P(x,y) such that x = (a cos^3 theta - 2) and y = (b sin^3 theta + 1) where a and b are constants and theta is parameter.

Answer 1

Given :

The x coordinate of the moving point P is given as :

$x = acos^3\theta -2$

The y coordinate of the point P is given as :

$y = bsin^3 \theta +1$

To Find :

Equation of the locus of the moving point P(x,y) is = ?

Solution :

Since in the  given equations a and b are constants and $\theta$ is parameter or variable , so we have to eliminate $\theta$ to get the locus :

So on arranging we can write as :

$acos^3\theta = x +2$   and $bsin^3\theta = y-1$

Or, $(acos^3\theta)^{\frac{2}{3}} = (x+2)^{\frac{2}{3}}$

Or, $a^{\frac{2}{3}} cos^2 =(x+2)^{\frac{2}{3}}$

Or, $cos^2\theta = \frac{(x+2)^{\frac{2}{3}}}{a^{\frac{2}{3}}}$    -(1)

Also :

$(bsin^3\theta)^{\frac{2}{3}}= (y-1)^{\frac{2}{3}}$

Or, $b^{\frac{2}{3}}sin^2\theta = (y-1)^{\frac{2}{3}}$

Or,$sin^2\theta = \frac{(y-1)^{\frac{2}{3}}}{b^{\frac{2}{3}}}$   - (2)

Now adding equations 1 and 2 :

$cos^2\theta + sin^2\theta = \frac{(x+2)^{\frac{2}{3}}}{a^{\frac{2}{3}}} + \frac{(y-1)^{\frac{2}{3}}}{b^{\frac{2}{3}}}$

Or,$\frac{(x+2)^{\frac{2}{3}}}{a^{\frac{2}{3}}} + \frac{(y-1)^{\frac{2}{3}}}{b^{\frac{2}{3}}} = 1$   (since $cos^2\theta + sin^2 \theta = 1$ )

Hence , the locus of the moving point P(x,y) is $\frac{(x+2)^{\frac{2}{3}}}{a^{\frac{2}{3}}} + \frac{(y-1)^{\frac{2}{3}}}{b^{\frac{2}{3}}} = 1$