Line passing through a point and normal to a curve

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  • #1
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Homework Statement


What is the equation of a straight line passing through (1,2) and normal to ##~x^2=4y##

Homework Equations


Slopes of perpendicular lines:
$$m_2=-\frac{1}{m_1}$$

The Attempt at a Solution


$$x^2=4y~\rightarrow~m_1=y'=\frac{x}{2}$$
$$m_2=-\frac{2}{x}$$
$$2=-\frac{2}{1}+b~\rightarrow~b=4$$
The equation of the line: ##~y=-2x+4##
The answer should be ##~x+y=3##
 

Answers and Replies

  • #2
FactChecker
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The x you use in m2 = -2/x is not the x value at the (1,2) point. It needs to be the x of the (x,y) values where the normal line intersects the equation.

In the given answer, the intersection is at (2,1) and that gives m2 = -1. That agrees with m1 = 1 = slope of y=x2/4 at (2,1). y' = x/2 = 1.
 
  • #3
scottdave
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If you use the point slope form of a line: y - y0 = m*(x - x0) then you can solve it. Assign (x0,y0) the point (1,2), and then (x,y) is a point on the parabola. Substitute y = x2 / 4, and then m = -2/x. You should then be able to solve for x.
 
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  • #4
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$$\frac{y_1-y_2}{x_1-x_2}=\frac{\frac{x^2}{4}-2}{x-1}=-\frac{2}{x}$$
$$\rightarrow~x^3=16~\rightarrow~x=2$$
$$\rightarrow~y=1$$
$$\rightarrow~y+x=3$$
Thank you Ray and Scott
 
  • #5
FactChecker
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You mean x3 = 8 → x=2.
 

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