Equations of normal
asked Nov 10, 2016 in Calculus Answers by anonymous

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(1) If y=(2x-1)/(1-x), y=(2x-1)(1-x)^-1, y'=2(1-x)^-1+(2x-1)(1-x)^-2=

2/(1-x)+(2x-1)/(1-x)^2=

(2-2x+2x-1)/(1-x)^2=1/(1-x)^2.

This is the slope of the tangent. The slope of the normal is -(1-x)^2=-1 when x=2.

Therefore the equation of the normal is y=-x+c where c is to be found by substitution of the coords when x=2.

At x=2, y=-3. Therefore -3=-2+c, c=-1. The equation of the normal is y=-x-1.

(2) P is found by solving -x-1=(2x-1)/(1-x); -(1+x)(1-x)=2x-1; x^2-1=2x-1, x^2-2x=0, x(x-2)=0 so x=0 or 2. We already know that x=2 is at the normal, so x=0 is the x-coord of P and the y-coord is -1, P(0,-1).

(3) y'=1/(1-x)^2 is the tangent. When x=0, y'=1 so the equation of the tangent is y=x+c and 0=-1+c, so c=1 and the equation of the tangent is y=x+1.

 

answered Nov 11, 2016 by Rod Top Rated User (417,500 points)
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