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Question

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθ, y = 1 + cosθ at θ = π2
(ii) x=2 at21+t2, y=2 at31+t2at t=12
(iii) x = at2, y = 2at at t = 1
(iv) x = asect, y = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi)
x = 3cosθ − cos3θ, y = 3sinθ − sin3θ [NCERT EXEMPLAR]

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Solution

(i)
x=θ+sinθ and y=1+cosθdxdθ=1+cosθ and dydθ=-sinθdydx=dydθdxdθ=-sinθ1+cosθSlope of tangent, m = dydxθ=π2=-sinπ21+cosπ2=-11+0=-1Now, x1, y1=π2+sinπ2, 1+cosπ2=π2+1, 1Equation of tangent is,y-y1=m x-x1y-1=-1x-π2-12y-2=-2x+π+22x+2y-π-4=0

Equation of normal is,y-y1=-1m x-x1y-1=1 x-π2-12y-2=2x-π-22x-2y=π

(ii) Equation of tangent:
x=2at21+t2 and y=2at31+t2dxdt=1+t24at-2at22t1+t22=4at1+t22 anddydt=1+t26at2-2at32t1+t22=6at2+2at41+t22dydx=dydtdxdt=6at2+2at41+t224at1+t22=6at2+2at44atSlope of tangent, m = dydxt=12=3a2+a82a=12a+a82a=13a8×12a=1316Now, x1, y1=2at21+t2,2at31+t2 =a21+14, a41+14=a254, a454=2a5, a5Equation of tangent is,y-y1=m x-x1y-a5=1316x-2a55y-a5=13165x-2a55y-a=13165x-2a80y-16a=65x-26a65x-80y-10a=013x-16y-2a=0

Equation of normal is,y-y1=-1m x-x1y-a5=-1613 x-2a55y-a5=-16135x-2a55y-a=-16135x-2a65y-13a=-80x+32a80x+65y-45a=016x+13y-9a=0

(iii)
x=at2 and y=2atdxdt=2at and dydt=2adydx=dydtdxdt=2a2at=1tSlope of tangent, m = dydxt=1=11=1Now, x1, y1=a, 2aEquation of tangent is,y-y1=m x-x1y-2a=1x-ay-2a=x-ax-y+a=0

Equation of normal:

Equation of normal is,y-y1=m x-x1y-2a=-1 x-ay-2a=-x+ax+y=3a

(iv)
x=a sec t and y=b tan tdxdt=a sec t tan t and dydt=b sec2tdydx=dydtdxdt=b sec2ta sec t tan t=bacosec tSlope of tangent, m = dydxt=t=bacosec tNow, x1, y1=a sec t, b tan tEquation of tangent is,y-y1=m x-x1y- b tan t=bacosec tx-a sec ty-b sin tcos t=ba sin tx-acos ty cos t-b sin tcos t=ba sin tx cos t-acos ty cos t-b sin t=ba sin tx cos t-aay sin t cos t-ab sin2t=bx cos t-abbx cos t-ay sin t cos t-ab1-sin2t=0bx cos t-ay sin t cos t=ab cos2tDividing by cos2t,bx sec t-ay tan t=ab

Equation of normal is,y-y1=m x-x1y- b tan t=-absin tx-a sec ty- b sin tcos t=-absin tx-acos ty cos t-b sin tcos t=-absin tx cos t-acos ty cos t-b sin t=-absin tx cos t-aby cos t-b2sin t=-ax sin t cos t+a2sin tax sin t cos t+by cos t=a2+b2sin tDividing both sides by sin t,ax cos t+by cot t=a2+b2

(v)
x=aθ+sinθ and y=a1-cosθdxdθ=a1+cosθ and dydθ=asinθdydx=dydθdxdθ=asinθa1+cosθ=sinθ1+cosθ=2sinθ2cosθ22cos2θ2=tanθ2 ...1Slope of tangent, m = dydxθ=tanθ2Now, x1, y1=aθ+sinθ, a1-cosθ Equation of tangent is,y-y1=m x-x1y-a1-cosθ=tanθ2x-aθ+sinθy-a2 sin2θ2=xtanθ2-aθtanθ2-atanθ2sinθy-a2 sin2θ2=xtanθ2-aθtanθ2-a2sinθ2cosθ22cos2θ22sinθ2cosθ2 (From (1))y-2a sin2θ2=x-aθtanθ2-2a sin2θ2y=x-aθtanθ2

Equation of normal is,y-a1-cosθ=-cotθ2x-aθ+sinθtan θ2y-a2 sin2θ2=-x+aθ+asinθtan θ2y-a2 1-cos2θ2=-x+aθ+asinθtan θ2y-2a+a 2 sin θ2 cos θ2=-x+aθ+asinθtan θ2y-2a+asinθ=-x+aθ+asinθtan θ2y-2a=-x+aθtan θ2y-2a+x-aθ=0

(vi)
x=3cosθ-cos3θ and y=3sinθ-sin3θdxdθ=-3sinθ+3cos2θsinθ=3sinθcos2θ-1=3sinθ-sin2θ=-3sin3θ and dydθ=3cosθ-3sin2θcosθ=3cosθ1-sin2θ=3cosθcos2θ=3cos3θdydx=3cos3θ-3sin3θ=-tan3θ ...1Slope of tangent, m = dydxθ=-tan3θNow, x1, y1=3cosθ-cos3θ, 3sinθ-sin3θ Equation of tangent is,y-y1=m x-x1y-3sinθ-sin3θ=-tan3θx-3cosθ-cos3θy-3sinθ-sin3θ=-sin3θcos3θx-3cosθ-cos3θycos3θ-3sinθcos3θ-sin3θcos3θ=-xsin3θ+3sin3θcosθ+sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθ+3sinθcos3θ+2sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθsin2θ+cos2θ+2sin3θcos3θxsin3θ+ycos3θ=3sin3θcosθ+2sin3θcos3θx+ycot3θ=3cosθ+2cos3θ

Equation of normal is,y-3cosθ-cos3θ=-1mx-3sinθ-sin3θy-3cosθ+cos3θ=-1tan3θx-3sinθ+sin3θy-3cosθ+cos3θ=-cos3θsin3θx-3sinθ+sin3θysin3θ-3sin3θcosθ+sin3θcos3θ=-xcos3θ+3cos3θsinθ-cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθ+3cos3θsinθ-2cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθsin2θ+cos2θ-2cos3θsin3θxcos3θ+ysin3θ=3sin3θcosθ-2cos3θsin3θxcot3θ+y=3cosθ-2cos3θ

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