Find the length arc of the parabola y2 = 4ax cut off by the line 3y = 8x.
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1
Answer:
a[√2+log(1+√2]
Step-by-step explanation:
Let A be the vertex and L an extremity of the Latus-Rectum so that at A,x=0 and at L,x=2a.
Now, y=
4a
x
2
so that
dx
dy
=
4a
1
.2x=
2a
x
∴arcAL=∫
0
2a
(1+(
dx
dy
)
2
)
dx
=∫
0
2a
(1+(
2a
x
)
2
)
dx
=
2a
1
∫
0
2a
(2a)
2
+x
2
dx
=
2a
1
⎣
⎢
⎡
2
x
(2a)
2
+x
2
+
2
(2a)
2
sinh
−1
(
2a
x
)
⎦
⎥
⎤
0
2a
=
2a
1
⎣
⎢
⎡
2
2a
(8a)
2
+2a
2
sinh
−1
1
⎦
⎥
⎤
Since sinh
−1
x=log(x+
1+x
2
) we have
Length of arc AL=a[
2
+log(1+
2
)]
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