Question

If the coordinates of a variable point P be $\left( {a\cos \theta ,b\sin \theta } \right)$ where $\theta $ is a variable quantity, find the locus of P.

Answer 1

Hint: By using the given coordinates of P $\left( {a\cos \theta ,b\sin \theta } \right)$ we get $\dfrac{x}{a} = \cos \theta $ and $\dfrac{y}{b} = \sin \theta $. Squaring and adding these we get the locus of the point P and we can see that the obtained locus is nothing but the general equation of the ellipse.

Complete step-by-step solution:
We are given that the coordinates of the point P to be $\left( {a\cos \theta ,b\sin \theta } \right)$
From this we get the $x$ coordinate of the point to be $a\cos \theta $
From which we get,
$x = a\cos \theta $
Dividing by $a$ on both sides we get
$\dfrac{x}{a} = \cos \theta $ ………… (1)
And we get the $y$ coordinate of the point to be \[b\sin \theta \]
From which we get,
$y = b\sin \theta $
Dividing by $b$ on both sides we get
$\dfrac{y}{b} = \sin \theta $ ………… (2)
Squaring and adding the equations (1) and (2) the following is obtained,
${\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = {\cos ^2}\theta + {\sin ^2}\theta $
We know the trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Substituting this identity within the above equation we get
${\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = 1$
We can see that this is a general equation of an ellipse.
Hence the locus of P is an ellipse.

Note: In geometry, a locus is one of a set of points which satisfies certain conditions. The results are usually a curve or a surface. For example, all points in a plane an equal distance from a centre point result in a circle. An ellipse is a plane curve surrounding the two focal points, such that for all points on that specific curve, the total of the two distances to the focal points is a constant. A circle is an ellipse, where both foci are at an equivalent point.