Tangents to parabola

  • Thread starter Mentallic
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  • #1
Mentallic
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Homework Statement



I am given the parabola [tex]y=\frac{x^2}{2}[/tex]

I need to find the equations of the 2 tangents to the parabola that pass through the point A(5,-2)



Homework Equations



[tex]y-y_{1}=m(x-x_{1})[/tex]

[tex]a = \frac{1}{2}[/tex]
therefore: tangents pass through the points [tex]P(p,\frac{p^{2}}{2})[/tex] and [tex]Q(q,\frac{q^{2}}{2})[/tex]

[tex]\frac{dy}{dx}=x[/tex]



The Attempt at a Solution



I began to check if the given point was outside the parabola
i.e. [tex]y_{1}<\frac{x^{2}_{1}}{2}[/tex]

[tex]-2<\frac{5}{2}[/tex] therefore, the point lies outside the parabola and there are 2 lines that will pass through the point, and are a tangent to the parabola.

[tex]y+2=m(x-5)[/tex] where there are 2 values of m, each intersecting the parabola only once. i.e. tangent to parabola.

From here I am totally stumped. I can't use the 1st derivative as I don't know the x value for which the gradient will pass through the point.
Any help would be much appreciated.
 
Last edited:

Answers and Replies

  • #2
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I need to find the equations of the 2 tangents to the parabola that pass through the point A(5,-2)
This means that your equation of your line must pass through two given points.

(5, -2) is one of them, so your equation [tex]y+2=m(x-5)[/tex] is a good place to start.

First, what is m? Remember, m is your slope...

Once you got that, now where does that second point have to be? If the line is tangent to the parabola with equation [tex] y=\frac{x^2}{2} [/tex]...
 
  • #3
HallsofIvy
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You want to find (x, y) such that y= m(x- 5)- 2 and y'(x)= m. You should certainly be able to find the derivative of y in terms of x. Putting that in for y' gives three equations for x, y, and m. this will, of course, reduce to a quadratic equation so you can expect two solutions: two tangents.
 
  • #4
Mentallic
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Thank you, I managed to solve it.
 
Last edited:

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