Question

A rod of fixed length ' l ' is sliding along the co-ordinate axes . Find the locus of the midpoint of the rod​

Answer 1

Answer:

$\huge\underline\mathfrak\red{ANSWER}$

Two rods of lengths a and b slide along coordinate axes such that their ends are concyclic. Locus of the center of the circle is

A

4(x

2

+y

2

)=a

2

+b

2

B

4(x

2

+y

2

)=a

2

−b

2

C

4(x

2

−y

2

)=a

2

−b

2

D

For the rod of length b

the extremes point (o, q) and

(o, q + b) and for the rod of

length a the extreme points

(p,o) and (p, p + a). The center of

circle will be intersection between

the ⊥ eq bisectors of rods

with length a and b.

(x,y)=(p+

2

a

,q+

2

b

)

So

x=p+

2

a

y=q+

2

b

x−p=

2

a

y−q=

2

b

__(1)

Now we need radius

But distance between (x, y) and

(p,o) and also distance between

(x,y) and (o,q) so

(x−p)

2

+y

2

=x

2

+(y−q)

2

__(2)

Now

(

2

a

)

2

+y

2

=x

2

+(

2

b

)

2

4

a

2

−b

2

=x

2

−y

2

a

2

−b

2

=4(x

2

−y

2

)