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October 29th, 2012, 08:14 AM  #1 
Member Joined: Oct 2012 Posts: 94 Thanks: 0  normal to the curve/normal to the circle
The line that is normal to the curve y = x^2 + 2x  3 at (1,0) intersects the curve at what other point? Find the points on the curve x^2 + xy + y^2 = 7 (a) where the tangent is parallel to the xaxis and (b) where the tangent is parallel to the y axis. For both of them, I differentiated. For the first, I found the slope and figured out the line. Don't know where to go from there, though. If someone could really walk me through this it would be appreciated. 
October 29th, 2012, 08:16 AM  #2 
Member Joined: Oct 2012 Posts: 94 Thanks: 0  Re: normal to the curve/normal to the circle
Oops, I wrote the wrong problem for the second one. The second problem should be: Find the two points where the curve x^2 + xy + y^2 = 7 crosses the x axis and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents?

October 29th, 2012, 11:45 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: normal to the curve/normal to the circle Quote:
Quote:
 
October 29th, 2012, 12:14 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: normal to the curve/normal to the circle
1.) We are given: Implicitly differentiate with respect to : This is the slope of the normal line, now us the pointslope formula to get the equation of the line: Now, substitute for in the equation of the parabola: Discard the known root and we have: and so Hence, the normal line also intersects the parabola at . 2.) We are given: To find where this curve crosses the xaxis, we set , and find: Thus, the two xintercepts are: Implicitly differentiating the given curve with respect to x, we find: Hence, we have shown that the common slope of the two tangents is 2. 
October 29th, 2012, 12:59 PM  #5 
Member Joined: Oct 2012 Posts: 94 Thanks: 0  Re: normal to the curve/normal to the circle
I just have one question...I guess more about the algebra. When I try to compute x1 = 4x^2  8x +12, I get 0 = 4x^2  9x + 13 I'm not exactly sure why, in your answer, all of the signs are opposite. 
October 29th, 2012, 01:05 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: normal to the curve/normal to the circle
The two are equivalent, just multiply through by 1, or equivalently, move everything to the other side.

October 29th, 2012, 01:10 PM  #7  
Member Joined: Oct 2012 Posts: 94 Thanks: 0  Re: normal to the curve/normal to the circle Quote:
Is that why dx/dy (2x + 2) = 1 becomes dx/dy = 1/2(x+1) as well? Thanks. Was a little confused about the negatives in that one too.  
October 29th, 2012, 01:11 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: normal to the curve/normal to the circle
Yes, exactly. We needed to solve for dx/dy to get the slope of the normal line. 

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