find equation of normal to curve y=log x at(e,1)
Answers
Answered by
2
Slope of tangent=dy/dx = 1/x
Slope of normal =-1/slope of tangent (They are perpendicular)
At e,1
Slope of tangent = 1/e
So slope of normal = -e
Using 2 slope point form
y-y1=m(x-x1)
y-1 = -e(x-e)
Or y = -ex + e^2 + 1
This is required equation
Mark brainliest if I deserve it
Slope of normal =-1/slope of tangent (They are perpendicular)
At e,1
Slope of tangent = 1/e
So slope of normal = -e
Using 2 slope point form
y-y1=m(x-x1)
y-1 = -e(x-e)
Or y = -ex + e^2 + 1
This is required equation
Mark brainliest if I deserve it
Similar questions