Grade 12Analytical Geometry

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

7 Years agoGrade 12

1 Answer

Ujjwal Singh

7 Years ago

x = a sin³t, y = a cos³t

dy / dt = -3a cos²t sint

dx / dt = 3a sin^2tt 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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1 Answer

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