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The slope of the tangent to the curve at a point (x,y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10, 9) on it is -3. The equation of the curve is Ans. y = (-3/16) (x-2) 2 + 3

The slope of the tangent to the curve at a point (x,y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10, 9) on it is -3. The equation of the curve is
Ans. y = (-3/16) (x-2)2 + 3
 

Grade:11

1 Answers

Arun
25763 Points
2 years ago
Dear student
 
 
the constant term should be 21
 
dy/dx=k(x-2)     where k is proportionality constant
 
at (10,9)  dy/dx=-3  hence
-3=k(10-2)   implies k=-3/8
 
dy/dx=(-3/8) *(x-2)
dy=(-3/8)*(x-2)*dx
 
integrating we get (-3/8)*((x-2)^2)/2)+c=y
 
now put x=10 and y=9 as the are the point of the curve and we get c=21
 
 
Regards
Arun (askIITians forum expert)

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