pore that x/a+y/b=1touches the curve y=be-x/a at the point wese curve crosses axis of y
Answer (1)
8th Dec, 2019
y=be^(-x/a)
dy/dx=-b/ae^(-x/a)
Now when this curve cuts y axis the value of x=0.
Hence the y value is b.Hence the curve cuts y axis at (0,b).The slope of this curve at (0,b) is -b/a.
Now we also have a straight line whose equation is given as y/b+x/a=1.
The straight line cuts y axis at (0,b).
Hence the straight line and the curve intersects at (0,b).
The slope of the straight line is -b/a.Hence the straight line touches the curve at (0,b).
dy/dx=-b/ae^(-x/a)
Now when this curve cuts y axis the value of x=0.
Hence the y value is b.Hence the curve cuts y axis at (0,b).The slope of this curve at (0,b) is -b/a.
Now we also have a straight line whose equation is given as y/b+x/a=1.
The straight line cuts y axis at (0,b).
Hence the straight line and the curve intersects at (0,b).
The slope of the straight line is -b/a.Hence the straight line touches the curve at (0,b).
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