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Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos2θ at
.
Asked by Deepak Last Modified
Answer x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)
dividing equations (2) by (1),
dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b sinθ/a
at θ = π/2 , dy/dx = 2bsinπ/2/a = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b
Slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal
x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)
dividing equations (2) by (1),
dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b
Harshit Bhardwaj
It is given that x = 1 − a sin θ and y = b cos2θ.
Therefore, the slope of the tangent at is given by,
Hence, the slope of the normal at
View 3 more Answers
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