Question

A rod of length 12 cm moves with its ends always touching the coordinate axes.
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

Answer 1

Answer:

Let AB be the rod making an angle θ with positive direction of x-axis and P(x,y) be the point on it such that AP=3cm

Now, PB=AB−AP=(12−3)cm=9cm (AB=12cm)

Draw PQ⊥OY and PR⊥OX

In △PBQ,

cosθ=

PB

PQ

=

9

x

In △PRA,

sinθ=

PA

PR

=

3

y

Since sin

2

θ+cos

2

θ=1

⇒(

3

y

)

2

+(

9

x

)

2

=1

⇒

81

x

2

+

9

y

2

=1

Thus the equation of the locus of point P on the rod is

81

x

2

+

9

y

2

=1

Answer 2

Let AB be the rod$~$

Where,

• A touches the x-axis

• B touches the y-axis

Let point P(x, y) Given that,

AB = Length of rod = 12 cm

AP = 3 cm

Now,

PB = AB − AP

= (12 − 3)cm

= 9cm (AB=12cm)

Now, Draw PQ⊥OY and PR⊥OX

In △PBQ,

$cosθ= \frac{PQ}{PB} = \frac{x}{9}$

In △PRA,

$sinθ= \frac{PR}{PA} = \frac{y}{3}$

Since sin²θ + cos²θ = 1

$⇒ (\frac{y}{3} )^{2} + (\frac{x}{9} )^{2} = 1$

$⇒ (\frac{y}{3} )^{2} + (\frac{x}{9} )^{2} = 1$

Hence the equation of the locus on the rod point P is

$(\frac{y}{3} )^{2} + (\frac{x}{9} )^{2} = 1$

$~$Thus locus of P is ellipse.