Question

find the equation of tangent and normal to the curve xy=10 at (2,5)

answer fastly☺......

Answer 1

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 ^_^ 

Answer 2

Answer:

Step-by-step explanation: