This lesson will cover a few simple examples involving locus.

**Q1.** Find the equation to the locus of a point (moving in a plane) such that:

(i) its distance from the point (3, 0) is three times the distance from the Y axis.

(ii) sum of the squares of its distances from the axes is equal to 4

(iii) its coordinates are (t^{2}, t + 1) where t is a parameter (this looks strange !)

**Solution. **(i) Just to remind you, what we need to do is find the relation between the coordinates of the variable point, which always holds true.

To start, we first write the given conditions in mathematical form. Let P(x, y) be the variable (or moving) point.

Then, its distance from (3, 0) is given by \(\sqrt{(x-3)^2+(y-0)^2}\)

And, its distance from the Y axis will be x (or -x, depending upon its location relative to the Y axis)

So, we have \(\sqrt{(x-3)^2+(y-0)^2}=\pm x\). After squaring and rearranging the terms, we get **y ^{2 }– 6x + 9 = 0**

(ii) Square of the distance from the Y axis is x^{2 }and square of the distance from the X axis is y^{2}

The sum is given to be 4. Therefore we have, **x ^{2 }+ y^{2 }= 4**.

Moving on…

(iii) What about this one? But where is the condition? Distance from this and distance from that? And ‘t’?

First we need to understand that why the point P should move, and secondly why under a condition.

Since ‘t’ is a parameter, with changing values of ‘t’, the x and y coordinates of P will change accordingly, making P a moving point.

But, since both x and y are related to the same variable ‘t’, they must be related to each other. This relation in fact will be the required equation to the locus of P.

Again, what we need to do is find a relation between x and y which always holds true. How to find it?

Think in this way: if x = t^{2}, and y = t +1, then how are x and y related?

Simple. Find the value of ‘t’ from one of the equations and substitute in the other.

From the second, we have t = y – 1, putting this value in the first we get **x = (y – 1)**^{2 }which is the required equation.

So, regardless of the value of t (because of which P moves), the relation x = (y – 1)^{2 }always holds true between x and y, which, by definition is the equation to the locus of P.

Don’t worry if you’re confused about the last question, focus on what we’re supposed to do when asked to find an equation, and things will be more clear.

You’ll come across a lot of questions involving the locus in the coming chapters, where I’ll provide you with a formal method to derive the equation to a locus.

The next couple of lessons will talk about something known as transformation of coordinates. Head on the next lesson to learn what it is. See you there !

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