Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

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 = π/2 . The curve is given as x = a sin3t , y = b cos3t Slope of the tangent = ππ¦/ππ₯ Here, ππ/ππ = (ππ/ππ)/(ππ/ππ) ππ/ππ = (π(π cos^3β‘γπ‘)γ)/ππ‘ = β3b cos^2 π‘ sinβ‘π‘ ππ/ππ = (π(π sin^3β‘γπ‘)γ)/ππ‘ = 3a sin^2β‘π‘ cosβ‘π‘ Hence, ππ¦/ππ₯ = (dy/dt)/(ππ₯/dt) = (β3ππππ ^2 π‘ sinβ‘π‘)/(3π sin^2β‘γπ‘ cosβ‘π‘ γ ) = (βπ πππβ‘π)/(π πππβ‘π ) Now, Slope of the tangent at "t = " π/2 is ππ/ππ = (βπ γcos γβ‘γπ/2γ)/(π γsin γβ‘γπ/2γ ) = (βπ(0))/(π(1)) = 0 To find Equation of tangent, we need to find point (x, y) Putting t = π/2 in equation of x and y π₯ = π sin3 (π/2) π=π π¦ = b cos3 (π/2) y = 0 Hence, point is (a, 0) Now, Equation of tangent at point (π, 0) and with slope 0 is y β 0 = 0 (x β π) y = 0