find the slope of the normal to the curve x=1-asinΘ y=bcos^2Θ at Θ=π/4
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It is given that x=1−asinθ and y=bcos
2
θ.
∴
dθ
dx
=−acosθ
and
dθ
dy
=2bcosθ(−sinθ)=−2bsinθcosθ
dx
dy
=
(
dθ
dx
)
(
dθ
dy
)
=
−acosθ
−2bsinθcosθ
=
a
2b
sinθ
Therefore, the slope of the tangent at θ=
2
π
is given by,
(
dx
dy
)
θ=
4
π
=
a
2b
Hence, the slope of the normal at θ=
2
π
is given by,
−
slope of the tangent atθ=
4
π
1
=
a
2b
−1
=−
2b
a
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