Finding equation of tangent/normal when point and curve is given

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Question 7 Find the equation of tangent to the curve given by x = a sin3 t , y = b cos3 t at a point where t = ๐/2 . The curve is given as x = a sin3t , y = b cos3t Slope of the tangent = ๐๐ฆ/๐๐ฅ Here, ๐๐/๐๐ = (๐๐/๐๐)/(๐๐/๐๐) ๐๐/๐๐ = (๐(๐ cos^3โกใ๐ก)ใ)/๐๐ก = โ3b cos^2 ๐ก sinโก๐ก ๐๐/๐๐ = (๐(๐ sin^3โกใ๐ก)ใ)/๐๐ก = 3a sin^2โก๐ก cosโก๐ก Hence, ๐๐ฆ/๐๐ฅ = (dy/dt)/(๐๐ฅ/dt) = (โ3๐๐๐๐ ^2 ๐ก sinโก๐ก)/(3๐ sin^2โกใ๐ก cosโก๐ก ใ ) = (โ๐ ๐๐๐โก๐)/(๐ ๐๐๐โก๐ ) Now, Slope of the tangent at "t = " ๐/2 is ๐๐/๐๐ = (โ๐ ใcos ใโกใ๐/2ใ)/(๐ ใsin ใโกใ๐/2ใ ) = (โ๐(0))/(๐(1)) = 0 To find Equation of tangent, we need to find point (x, y) Putting t = ๐/2 in equation of x and y ๐ฅ = ๐ sin3 (๐/2) ๐=๐ ๐ฆ = b cos3 (๐/2) y = 0 Hence, point is (a, 0) Now, Equation of tangent at point (๐, 0) and with slope 0 is y โ 0 = 0 (x โ ๐) y = 0