1 answer

Find the equation of tangent to the curve #y=2/(2x+9)# and passing through #(3,0)#. What is the equation of tangent to

Question:

Find the equation of tangent to the curve #y=2/(2x+9)# and passing through #(3,0)#. What is the equation of tangent to the curve at #x=x_0#?

Answers

Please see below.

Explanation:

Let the tangent passing through #(3,0)# have a slope of #m#. Hence its equation would be #y=m(x-3)=mx-3m#.

And line cuts the curve #y=2/(2x+9)# at points given by

#mx-3m=2/(2x+9)# or #2mx^2-6mx+9mx-27m=2#

or #2mx^2+3mx-27m-2=0#

As tangent cuts the curve only at one point, we must have discriminant of the equation #0# i.e.

#(3m)^2-4xx2mxx(-27m-2)=0#

i.e. #9m^2+216m^2+16m=0# i.e. #m(225m+16)=0#

Hence, either #m=0# or #m=-16/225# i.e. slopes of tangency are #0# and #-16/225#;

and tangents are #y=0# and #y=-16/225(x-3)# or #225y+16x-48=0#.

Note from the following graph that #y=0# is the tangent to the curve at #(+-oo,0)#.

graph{y(y-2/(2x+9))(225y+16x-48)=0 [-10, 10, -5, 5]}

The slope of tangent to a curve #y=f(x)# is given by #f'(x)# i.e. #(df)/(dx)#. As #y=f(x)=2/(2x+9)#

#f'(x)=2xx(-1)/(2x+9)^2xx2=-4/(2x+9)^2#

and at #x=x_0# slope #k# is given by #k=-4/(2x_0+9)^2# and tangent at #(x_0,2/(2x_0+9))# is

#(y-2/(2x_0+9))=-4/(2x_0+9)^2(x-x_0)#

.

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