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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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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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