How do you find the equation of the tangent and normal line to the curve #y=(2x+3)/(3x-2)# at #(1,5)#?

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Answer 1 · 2017-01-14T10:09:52

#(1)"Eqn. of Tgt. : "13x+y-18=0#.

#(2)"Eqn. of Normal : "x-13y+64=0#.

Recall that #dy/dx# gives the slope of tangent (tgt.) to the curve at

the general point #(x,y)#.

Now, #y=(2x+3)/(3x-2)={2/3(3x-2)+13/3}/(3x-2)=2/3+13/3(3x-2)^-1#

#rArr dy/dx=0+13/3{-1(3x-2)^(-1-1)d/dx(3x-2)}#

#:.dy/dx=-13/(3x-2)^2#.

#:." The Slope of the Tgt. at the Point "(1,5)" is, "-13.#

Also, the tgt. passes through #(1,5)#.

Therefore, by the Slope-Pt. Form , the eqn. of tgt. is,

#y-5=-13(x-1), i.e., 13x+y-18=0#.

As regards the eqn. of the Normal, it is #bot# to tgt. at #(1,5)#.

So, the eqn. of normal is # : y-5=1/13(x-1), or, x-13y+64=0#.

Enjoy Maths.!