Expert: Paul Klarreich - 11/21/2011

Question
the normal to the curve y= x^2-9x at point P is parallel to the straight line y= -x+8. find the equation of the normal to the curve at point P.
answer:y= -x-15
can you show me the full working done for this question.

Answer
Questioner: Quaternion
Country: Sabah, Malaysia
Category: Calculus
Private: No
Subject: Differentiation: equation to normal
Question: the normal to the curve y= x^2-9x at point P is parallel to the straight line y= -x+8. find the equation of the normal to the curve at point P.
answer:y= -x-15
can you show me the full working done for this question.
...........................................
Yes, but you will fill in the details:

y = -x + 8  has a slope of  -1.  

So your 'normal line' has a slope of -1, and the tangent line has slope equal to  +1.  (negative reciprocal)

Now you will:

A. Find dy/dx for the curve  y = x^2 - 9x

B. Find x  where  dy/dx = +1.  Call this  x0.

C. Find y at that point.  (The point is ON the curve.)  Call this y0.

D. Use  m = -1  (for the normal),  x0, and y0,  and the point-slope form:

y - y0 = m(x - x0)

and you have your equation.