Line passing through a point and normal to a curve

[Karol]

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 $~x+y=3$

 

[FactChecker]

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.

 

[scottdave]

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.

 

[Karol]

$$\frac{y_1-y_2}{x_1-x_2}=\frac{\frac{x^2}{4}-2}{x-1}=-\frac{2}{x}$$
$$\rightarrow~x^3=16~\rightarrow~x=2$$
$$\rightarrow~y=1$$
$$\rightarrow~y+x=3$$
Thank you Ray and Scott

 

[FactChecker]

You mean x³ = 8 → x=2.