Find the slope of the normal to the curve x = 1 − a sin θ , y = b cos 2 θ at .
The given curve is defined as,
x = 1 − a sin θ y = a cos 2 θ
Slope of the normal to the curve is given as,
Slope of the normal at θ = θ 0 = − 1 Slope of the tangent at θ = θ 0 (1)
Slope of tangent to a curve y at a given point θ = θ 0 is given by,
Slope = ( d y d x ) θ = θ 0
The given curve is in parametric form hence in order to find ( d y d x ) θ = θ 0 we need to find ( d x d θ ) θ = θ 0 and ( d y d θ ) θ = θ 0 such that,
( d y d x ) θ = θ 0 = ( d y d θ ) θ = θ 0 ( d x d θ ) θ = θ 0 (2)
The derivative ( d x d θ ) is given by,
( d x d θ ) = [ − a cos θ ]
The derivative ( d y d θ ) is given by,
( d y d θ ) = − 2 b sin θ ⋅ cos θ
Substitute − 2 b sin θ ⋅ cos θ for ( d y d θ ) and [ - a cos θ ] for ( d x d θ ) at θ = π 2 in equation (2).
( d y d x ) θ = π 2 = ( − 2 b sin θ ⋅ cos θ − a cos θ ) θ = π 2 = ( 2 b a sin θ ) θ = π 2 = 2 b a
Hence, the slope of normal at θ = π 2 from equation (1) is given as,
− 1 2 b a = − a 2 b
The given curve is defined as,
Slope of the normal to the curve is given as,
Slope of tangent to a curve
The given curve is in parametric form hence in order to find
The derivative
The derivative
Substitute
Hence, the slope of normal at
