The parametric equations of a curve are x=cosθ and y=1+sinθ. Show that the curve is a circle with centre(0,1) and radius 1 unit. Find the equation of the chord at the points on the curve where θ=#π/6# and θ=#π/3# respectively. ?

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Answer 1 · 2018-06-25T15:26:12

# 2(x+y)-(3+sqrt3)=0#.

#x=costheta, y=1+sintheta#.

#:. x=costheta, (y-1)=sintheta#.

But, #cos^2theta+sin^2theta=1#.

#:. x^2+(y-1)^2=1, i.e., #

# (x-0)^2+(y-1)^2=1^2#.

Knowing that, #(x-h)^2+(y-k)^2=r^2# represents a circle with

centre at the point #(h,k)# & radius #r#, we conclude

that, the given parametric eqns. describes a circle with centre

at #(0,1)# and radius #1#.

Let #P(theta)=P(pi/6)=(costheta,1+sintheta)=(cos(pi/6),1+sin(pi/6))#

#:. P(pi/6)=(sqrt3/2,1+1/2)=(sqrt3/2,3/2)#.

Likewise, #Q(pi/3)=(1/2,1+sqrt3/2)#.

#:."The slope of "PQ={(1+1/2)-(1+sqrt3/2)}/(sqrt3/2-1/2)=-1#

Using this slope and the point #P#, we get, the eqn. of the chord #PQ#

#y-3/2=-1(x-sqrt3/2), or, 2(x+y)-(3+sqrt3)=0#.