Finding parametric equation of the line tangent to the parabola

[shemer77]

Homework Statement

Find a parametric equation of the line that satisfies the condition:
The line that is tangent to the parabola y=x^2 at the point (-2,4)

The Attempt at a Solution

My answer came out to
<x,y> = <-2,4> + t<1,2>

 

[glebovg]

Any curve whose equation is given by y = f(x) can be parametrized as
r = <x, f(x)>.

 

[dynamicsolo]

For a parametric curve, the slope at a point (x₀, y₀) is

$$(\frac{dy}{dx}) \vert_{(x_{0},y_{0})} = \frac{dy/dt}{dx/dt} \vert_{(x_{0},y_{0})} .$$

What is dy/dx for the parabola at (-2, 4)? What is the slope of your line? (There is a discrepancy.) How did you work out your answer?glebovg, shemer77 has written the pair of equations as a single vector equation -- that notation is acceptable for a system of equations.

 

[shemer77]

sorry, I am not really sure what your asking? What I did was I took the derivative of y=x^2 b/c that would give me the slope of a line tangent to the parobala then i used that slope as my vector. Are you asking what's the derivative at (-2,4)?

 

[HallsofIvy]

You have told us, twice, what you got without telling us how you got it. The answer you got is NOT correct but since you haven't shown how you got that, we can't tell what you did wrong.

 

[shemer77]

I mentioned it in my last post

What I did was I took the derivative of y=x^2 b/c that would give me the slope of a line tangent to the parobala then i used that slope as my vector.

Are you saying I should take the derivative which would give me dy/dx=2x and thus dy/dx=-4? and would I put that into my original equation? <x,y> = <-2,4> + t<1,-4>

 

[flyingpig]

My answer looked completely different from yours and based on what HallsofIvy says, I think I am not wrong.

What I did was do the line the normal way, that is

Find y = y'(x - x_0) + y_0

Then let x = t

 

[HallsofIvy]

I mentioned it in my last post



Are you saying I should take the derivative which would give me dy/dx=2x and thus dy/dx=-4? and would I put that into my original equation? <x,y> = <-2,4> + t<1,-4>

Just saying you "took the derivative" does not tell us how you got that equation! The "algebraic" form of the equation would be y= m(x-1)- 4 where m is the value of the derivative at (1, -4). One possible parametric equation for that would be t<1, m>+ <1, -4>. Do you see why?

 

[shemer77]

Hmm, I think so but I don't see how my answer of
<x,y> = <-2,4> + t<1,-4>
is wrong?

 

[flyingpig]

Just saying you "took the derivative" does not tell us how you got that equation! The "algebraic" form of the equation would be y= m(x-1)- 4 where m is the value of the derivative at (1, -4). One possible parametric equation for that would be t<1, m>+ <1, -4>. Do you see why?

That's odd I got q(t) = <0,-4> + t<1,m>