My Math Forum normal to the curve/normal to the circle

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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 x-axis 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
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Re: normal to the curve/normal to the circle

Quote:
 Originally Posted by cheyb93 The line that is normal to the curve y = x^2 + 2x - 3 at (1,0) intersects the curve at what other point?
You say you have found the equation of the normal line, in the form y= mx+ b, I presume. To determine whete that line itersects the curve, set them equal, $x^2+ 2x- 3= mx- b$ or [/latex]x^2+ (2- m)+ b- 3= 0[/tex]. You already know one solution so it should be easy to find the other..

Quote:
 Find the points on the curve x^2 + xy + y^2 = 7 (a) where the tangent is parallel to the x-axis 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.
Any line parallel to the x- axis is of the form "y= constant", which has slope 0. Any line parallel to the x- axis is of the form "x= constant", which does not have a "slope".

 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: 521 Math Focus: Calculus/ODEs Re: normal to the curve/normal to the circle 1.) We are given: $y=x^2+2x-3$ Implicitly differentiate with respect to $y$: $1=2x\frac{dx}{dy}+2\frac{dx}{dy}=\frac{dx}{dy}(2x+ 2)$ $-\frac{dx}{dy}=-\frac{1}{2(x+1)}$ $-\frac{dx}{dy}\|_{x=1}=-\frac{1}{2(1+1)}=-\frac{1}{4}$ This is the slope of the normal line, now us the point-slope formula to get the equation of the line: $y-0=-\frac{1}{4}(x-1)$ $y=-\frac{1}{4}x+\frac{1}{4}$ Now, substitute for $y$ in the equation of the parabola: $-\frac{1}{4}(x-1)=x^2+2x-3$ $x-1=-4x^2-8x+12$ $4x^2+9x-13=0$ $(4x+13)(x-1)=0$ Discard the known root $x=1$ and we have: $x=-\frac{13}{4}$ and so $y=-\frac{1}{4}$$-\frac{13}{4}-1$$=\frac{17}{16}$ Hence, the normal line also intersects the parabola at $$$-\frac{13}{4},\frac{17}{16}$$$. 2.) We are given: $x^2+xy+y^2=7$ To find where this curve crosses the x-axis, we set $y=0$, and find: $x^2=7$ $x=\pm\sqrt{7}$ Thus, the two x-intercepts are: $$$-\sqrt{7},0$$,\,$$\sqrt{7},0$$$ Implicitly differentiating the given curve with respect to x, we find: $2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0$ $\frac{dy}{dx}$$x+2y$$=-(2x+y)$ $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$ $\frac{dy}{dx}\|_{y=0}=-\frac{2x+0}{x+2(0)}=-2$ 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 x-1 = -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: 521 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
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Re: normal to the curve/normal to the circle

Quote:
 Originally Posted by MarkFL The two are equivalent, just multiply through by -1, or equivalently, move everything to the other side.

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: 521 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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