Find the equation of the tangent to the curve y = 50/(2x-1)^2 at point (3,2), in the form ax+by+c = 0.
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Given the curve y= 50/(2x-1)^2
We need to find the tangent line at the point (3,2)
We know that the equation of the tangent line is:
(y-y1) = m(x-x1) where (x1,y1) is any point on the line and m is the slope.
We know that the tangent line passes through the point (3,2)
==> (y-2) = m(x-3)
Now we will calculate the slope.
The slope of the tangent line is the derivative of the curve at the point x= 3
==> y' = [50(2x-1)^-2]'
= 50*-2*2 ( 2x-1)^-3 = - 200/ (2x-1)^3
Now we will substitute with x= 3
==> m = -200 / (5^3) = -200 / 125 = -8/5
Then we will substitute with the slope m = -8/5
==> (y -2) = (-8/5) ( x-3)
==> y-2 = (-8/5)x + 24/5
We will multiply by 5.
==> 5y - 10 = -8x + 24
Now we will move all terms to the left side.
==> 8x + 5y - 34 = 0
Then the equation of the tangent line is 8x + 5y - 34 = 0
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