Find the equations of the tangent and normal to the curve x = a sin³ θ and y = a cos³θ at θ = π
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Answered by
1
Correct option is
C
x=y
Slope of normal to the curve = −
(
dx
dy
)
θ=π/4
1
x=acos
3
θ,y=asin
3
θ
dθ
dx
=3acos
2
θ(−sinθ),
dθ
dy
=3asin
2
θcosθ
∴
dx
dy
=
dx/dθ
dy/dθ
=
−3acos
2
θsinθ
+3asin
2
θcosθ
=−tanθ
∴(
dx
dy
)
θ=π/4
=−1
∴ Equation of normal =(y−y
0
)=
(
dx
dy
)
θ=π/4
−1
[x−x
0
]
y−
2
2
a
=
−1
−1
(x−
2
2
a
) [x
0
=acos
3
4
π
=
2
2
a
y−
2
2
a
=x−
2
2
a
y
0
=asin
3
4
π
=
2
2
a
]
y=x
only 1 brainlist left
Answered by
0
Answer:
Th.e equ.ation of a pla.ne pas.sing thr.ough (1,−2,3) is a(x−1)+b(y+2)+c(z−3)=0 .(i)
It pas.ses thro.ugh (−1,2,−1) an.d is para.llel to the giv.en lin.e
=−1c
He.nce, a:b:c=2:0:−1.
hope help uh ❤
.
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