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if ot and on are perpendicular dropped from the origin to the tangent and normal to the curve x=asin3t, y=acos3t at an arbitary point, then 4OT2+ON2=a2?

Aman Bhardwaj , 5 Years ago
Grade 12
anser 1 Answers
Ujjwal Singh

Last Activity: 5 Years ago

x = a sin3t, y = a cos3t
 
dy / dt = -3a cos2t sint
 
dx / dt = 3a sin2tt cost
 
dy / dx = (dy/dt) / (dx/dt)
            = −cot t
dx / dy = tan t
 
  • Eq. of tangent at point t –
y – acos3t = – cot t (x – asin3t), which simplifies to –
x + y tant – a sint = 0
 
Length of OT =  \dfrac{|(0) + (0)tant - asint|}{\sqrt{1 + tan^{2}t}}
                      = |asin2t| / 2
 
  • Eq. of normal at point t –

    y – acos3t = tan t (x – asin3t), which simplifies to –
    x sint – y cost + a cos2t = 0

    Lenght of ON = \dfrac{|(0)sint - (0)cost + acos2t|}{\sqrt{sin^{2}t + cos^{2}t}}
                          = |acos2t|
 
Therefore, 4OT2 + ON2 = a2 (sin22t + cos22t)
                                       = a2
 

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