 Hello and welcome to the session. Let us discuss the following question. It says, a rod of length 12 cm moves with its ends always touching the coordinate axis. Determine the equation of the locus of a point P on the rod which is 3 cm from the end in contact with the x-axis. Let us now move on to the solution. Here we have to find the locus of the point P on the rod which is 3 cm from the end in contact with x-axis. For that we need to understand what is a locus. Locus is a moving point which satisfies certain condition. So we have a rod of length 12 cm. So let AB be the rod making an angle theta with OX. Now let XY be the point at the rod such that BP is 3 cm. So we have to find locus of this point. Now AB is 12 cm and BP is 3 cm. So AB is equal to AB minus BP that is 12 minus 3 cm that is 9 cm. Now we draw perpendicular from P or PQ and PR on the coordinate axis. Now in triangle AQ cos theta is equal to base upon hypotenuse base is QP hypotenuse is AB. Now QP has length X and AP is 9 cm. So cos theta is equal to X upon 9. Now in triangle PRB sin theta is equal to PR upon PB or BP. Now PR has length Y cm because the coordinate of the point P is PXY. So Y is the length of PR and PB has length 3 cm. Now we know that cos square theta plus sin square theta is equal to 1. So X upon 9 square plus Y upon 3 square is equal to 1 which is equal to X square upon 81 plus Y square upon 9 is equal to 1. So this is the locus of the point P and we see that it is an ellipse. This is the point P phrases an ellipse. So the locus of the point P is an ellipse given by X square upon 81 plus Y square upon 9 is equal to 1. And this completes the question and the session. Bye for now. Take care. Have a good day.