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.!

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