find the equation of tangent and normal to the curve xy=10 at (2,5)
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equation of tangent is given by = (y-y1)= M(x-x1)
M=dy/dx of the curve
xy=10
y=10/x
dy/dx = (-10/x^2 ) at x = 2
M= -10/4=-5/2 so we get the slope now
y1=5 x1 =2
y-y1=m(x-x1) be equation of tangent
y-5=-5/2(x-2)
y+5/2x -10 = 0
2y + 5x - 20 =0 is the equation of tangent
now for equation of normal
we need Mn by using this equation
M*Mn = -1
Mn *-5/2 =-1
Mn =2/5 for equation of tangent
y-y1 =Mn (x-x1)
y-5 = 2/5(x-2)
y-2/5 x -5 + 4/5 = 0
5y - 2x -21 = 0 be the equation of tangent hope u get it ask any doubt if u have in comment section mark it as brainliest one please i want to increase my rank i just did this for u ^_^
M=dy/dx of the curve
xy=10
y=10/x
dy/dx = (-10/x^2 ) at x = 2
M= -10/4=-5/2 so we get the slope now
y1=5 x1 =2
y-y1=m(x-x1) be equation of tangent
y-5=-5/2(x-2)
y+5/2x -10 = 0
2y + 5x - 20 =0 is the equation of tangent
now for equation of normal
we need Mn by using this equation
M*Mn = -1
Mn *-5/2 =-1
Mn =2/5 for equation of tangent
y-y1 =Mn (x-x1)
y-5 = 2/5(x-2)
y-2/5 x -5 + 4/5 = 0
5y - 2x -21 = 0 be the equation of tangent hope u get it ask any doubt if u have in comment section mark it as brainliest one please i want to increase my rank i just did this for u ^_^
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