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ADNAN MUHAMMED

Grade 12,

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prove that the straight line x/a+y/b=1 touches the curve y=be^-x/a at that point where the curve cuts the y-axis also find the equation of normal at the same point

prove that the straight line x/a+y/b=1 touches the curve y=be^-x/a at that point where the curve cuts the y-axis also find the equation of normal at the same point 
 

Grade:12th pass

1 Answers

yathartha gupta
71 Points
6 years ago
Straight line touches the curve at y axis so there x will be 0Put the value in equationy=be^-0/ay=be^0Y=b.So the co-ordinates of the point will be(0,b)By diff. the curve equationY=be^-x/ady/dx=-b/ae^-x/a[dy/dx](0,b) = -b/ae^0 = -b/aSo the slope of normal will be a/bThe equation of normal at (0,b) will bey-y1=m(x-x1)y - b = a/b(x - 0)y - b = ax/by - ax/b = bby - ax =b^2..................Ans.

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