Find the equations of tangent and mormal to the curve xy=10at(2,5)
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y=10/x
differentiate wrt x
dy/dx= -10/x^2= slope
substitute x= 2 the slope = -10/4 = -5/2
then tangent passing through (2,5) with slope -5/2 is
(y-5)=-5/2(x-2)
for normal multiply slope with -1 and reciprocal ie 2/5
the equation of normal is
(y-5)=2/5(x-2)
differentiate wrt x
dy/dx= -10/x^2= slope
substitute x= 2 the slope = -10/4 = -5/2
then tangent passing through (2,5) with slope -5/2 is
(y-5)=-5/2(x-2)
for normal multiply slope with -1 and reciprocal ie 2/5
the equation of normal is
(y-5)=2/5(x-2)
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