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 ${\displaystyle \frac{x}{a}}=\cos \theta$ and ${\displaystyle \frac{y}{b}}=\sin \theta$. 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
${\displaystyle \frac{x}{a}}=\cos \theta$ ………… (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
${\displaystyle \frac{y}{b}}=\sin \theta$ ………… (2)
Squaring and adding the equations (1) and (2) the following is obtained,
${\left({\displaystyle \frac{x}{a}}\right)}^2+{\left({\displaystyle \frac{y}{b}}\right)}^2={\cos }^2\theta +{\sin }^2\theta$
We know the trigonometric identity ${\cos }^2\theta +{\sin }^2\theta =1$
Substituting this identity within the above equation we get
${\left({\displaystyle \frac{x}{a}}\right)}^2+{\left({\displaystyle \frac{y}{b}}\right)}^2=1$
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.