To find the length of the tangent at the point P(‌
Ï€
4
) on the curve x2∕3+y2∕3=22∕3, we can proceed as follows: The given equation of the curve can be parameterized as: x=2cos3θ,‌‌y=2sin‌3θ For the point P, we substitute θ=‌
Ï€
4
: x=2cos3‌
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4
,‌‌y=2sin‌3‌
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4
Calculating the coordinates: x=2(‌
1
√2
)3=‌
2
2√2
=‌
1
√2
,‌‌y=2(‌
1
√2
)3=‌
2
2√2
=‌
1
√2
So the point is (‌
1
√2
,‌
1
√2
). Next, differentiate the curve equation x2∕3+y2∕3=22∕3 with respect to x : ‌
2
3
x−1∕3+‌
2
3
y−1∕3‌
dy
dx
=0 Solving for ‌
dy
dx
: ‌
dy
dx
=−(‌
y
x
)1∕3 At the point (‌
1
√2
,‌
1
√2
), ‌
dy
dx
=−1 The formula to find the length of the tangent at a point is: L=y√1+(‌
dx
dy
)2 Given ‌
dx
dy
=−1, the length of the tangent at the point is calculated as: L=‌