Question

A rod of length 12cm moves with its ends always touching the coordinate axis. Determine the equation of a locus of a point P on the rod on the rod, which is 3 cm from the end in contact with the x-axis.

Answer 1

Hint: Firstly draw the diagram on a coordinate axis with given parameters or values and carefully assign the variables to it and its corresponding values and then try to proceed. We have to know the properties when two parallel lines are cut by a transversal and basic trigonometric functions.

Complete step-by-step solution:
Let AB be the rod where A touches the x-axis and B touches the y-axis

Let point \[P(x,y)\]
Given AB= length of rod=12cm and AP=3cm
PB=AB-AP
PB=12-3=9cm
Drawing PQ perpendicular to BO and PR perpendicular to OA
PQ=x and PR=y
Let $\angle {PAR} $ = $\theta$
Now, PQ and AO are parallel lines(as both are perpendicular to y-axis) and BA is transversal
So, $\angle{BPQ} $ = $\angle{PAR}$ =$\theta$
In the right angled triangle BPQ
We know that the formula of \[\cos \theta \] is given by
\[\cos \theta =\dfrac{base}{hypotenuse}\]
\[\cos \theta =\dfrac{PQ}{BP}\]
\[\cos \theta =\dfrac{x}{9}\]. . . . . . . . . . . . . . . . . . . . . . . . (1)
In the right angled triangle PAR
We know that the formula of \[\sin \theta \] is given by
\[\sin \theta =\dfrac{height}{hypotenuse}\]
\[\sin \theta =\dfrac{PR}{AP}\]
\[\sin \theta =\dfrac{y}{3}\] . . . . . . . . . . . . . . . . . . . . . . . (2)
We know the trigonometric identity that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Now putting \[\sin \theta =\dfrac{y}{3}\] and \[\cos \theta =\dfrac{x}{9}\] we will get
\[{{\left( \dfrac{y}{3} \right)}^{2}}+{{\left( \dfrac{x}{9} \right)}^{2}}=1\]
\[\dfrac{{{x}^{2}}}{81}+\dfrac{{{y}^{2}}}{9}=1\]
So, the locus of a point P on the rod which is 3 cm from end in contact with x-axis is \[\dfrac{{{x}^{2}}}{81}+\dfrac{{{y}^{2}}}{9}=1\]

Note: We have to know the basic trigonometric identity that is \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and we have to note that the equation of locus of a point p satisfies the equation of the ellipse as the general form of ellipse is \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. So the locus of a point P is an ellipse. Note that the set of all points which satisfy the given geometrical conditions is called locus.