# 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?

Ujjwal Singh
13 Points
3 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