To find the length of the tangent at the point
P() 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=2sin3θFor the point
P, we substitute
θ= :
x=2cos3,y=2sin3Calculating the coordinates:
x=2()3==,y=2()3==So the point is
(,).
Next, differentiate the curve equation
x2∕3+y2∕3=22∕3 with respect to
x :
x−1∕3+y−1∕3=0Solving for
:
=−()1∕3At the point
(,),
=−1The formula to find the length of the tangent at a point is:
L=y√1+()2Given
=−1, the length of the tangent at the point is calculated as:
L=√1+(−1)2=×√2=1Thus, the length of the tangent is 1 .