Equations of two Lines that are Tangent to a Parabola

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Drakkith
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Homework Statement


[/B]
Find equations of both lines through the point (2,-3) that are tangent to the parabola y=x2+x.

Homework Equations


Slope formula: (Y2-Y1)/(X2-X1) = M

The Attempt at a Solution



Here's what I think I need to do. First I think I need to find the derivative of the parabola. Then I need to use the point (2,-3) as point 2 in the slope equation Y2-Y1/X2-X1 to find two X values that give me a slope equal to the derivative at that x-value.

Derivative: y'=2x+1

Finding slope: The tangent line passes through (2,-3) and the parabola. Since the slope of the tangent line is found by the derivative, that means that y2-y1/x2-x1 should be equal to the derivative 2x+1, or: [-3-(2x+1)]/(2-x) = 2x+1

So that becomes: (-3-2x-1)/(2-x) = 2x+1, then (-4-2x)/(2-x) = 2x+1.
Moving the bottom over: -4-2x = (2x+1)(2-x)
Foiling: -4-2x = 4x-2x2+2-x.
Combining: -4-2x = -2x2+3x+2
Moving the right side over and combining: 2x2-5x-6 = 0
So the tangent lines should be located at the x-coordinates found by finding x in the above quadratic, right?
 

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  • #2
Orodruin
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I think you are over-complicating things. A line through a given point has one parameter, which can be taken to be the slope. Solving for a line intersecting a parabola is equivalent to solving for the roots of a second order polynomial. For the lines which are tangent to the parabola, you should get only one (double) root. Vary the parameter of the line in such a way that this is fulfilled.
 
  • #3
SteamKing
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Homework Statement


[/B]
Find equations of both lines through the point (2,-3) that are tangent to the parabola y=x2+x.

Homework Equations


Slope formula: (Y2-Y1)/(X2-X1) = M

The Attempt at a Solution



Here's what I think I need to do. First I think I need to find the derivative of the parabola. Then I need to use the point (2,-3) as point 2 in the slope equation Y2-Y1/X2-X1 to find two X values that give me a slope equal to the derivative at that x-value.

Derivative: y'=2x+1

Finding slope: The tangent line passes through (2,-3) and the parabola. Since the slope of the tangent line is found by the derivative, that means that y2-y1/x2-x1 should be equal to the derivative 2x+1, or: [-3-(2x+1)]/(2-x) = 2x+1

So that becomes: (-3-2x-1)/(2-x) = 2x+1, then (-4-2x)/(2-x) = 2x+1.
Moving the bottom over: -4-2x = (2x+1)(2-x)
Foiling: -4-2x = 4x-2x2+2-x.
Combining: -4-2x = -2x2+3x+2
Moving the right side over and combining: 2x2-5x-6 = 0
So the tangent lines should be located at the x-coordinates found by finding x in the above quadratic, right?

Before you go all mathy on this problem, you should make a quick sketch of the parabola and plot the given point (2, -3). The slopes of the tangent lines could be quite different.
 
  • #4
Drakkith
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For the lines which are tangent to the parabola, you should get only one (double) root. Vary the parameter of the line in such a way that this is fulfilled.

I'm sorry I don't understand what you're telling me to do.

Edit: Let me think on this some more.
 
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  • #5
Drakkith
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Before you go all mathy on this problem, you should make a quick sketch of the parabola and plot the given point (2, -3). The slopes of the tangent lines could be quite different.

Hmm, you mean the two lines will have different slopes? If so, then that is exactly what I was expecting. Can you elaborate on how sketching this helps me?
 
  • #6
SteamKing
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Hmm, you mean the two lines will have different slopes? If so, then that is exactly what I was expecting. Can you elaborate on how sketching this helps me?
Well for one thing, a sketch could help you confirm whether the slopes of your two tangent lines are correct. Another thing is that you can also see approximately where the tangent points themselves are located on the parabola.
 
  • #7
Drakkith
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Well for one thing, a sketch could help you confirm whether the slopes of your two tangent lines are correct. Another thing is that you can also see approximately where the tangent points themselves are located on the parabola.

You make a valid point. :biggrin:
 
  • #8
Orodruin
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I'm sorry I don't understand what you're telling me to do.

Edit: Let me think on this some more.
You have two functions, the parabola f(x) and the line g(x,k), where k is the slope. Fixing k, f(x)-g(x,k) is a second order polynomial and its roots are where f and g are equal. Since it is a second order polynomial, it has two roots. If the roots are real and different, the line goes through the parabola twice and is never tangent (since the parabola is convex). If the roots are imaginary, the line never meets the parabola (for real x). If the roots are a double root, the line only meets the parabola once and must be tangent to it.

You can look at it this way: let us say you start with a line such that you have two points of intersection. You will be able to change the slope in such a direction that the x-coordinates of the intersection points come closer. Continuing to do this you will eventually reach the point where they meet, this is the point where the line is tangent to the parabola. Continuing to change the slope from this point on will result in the line and parabola never meeting.

The exception to all of this is when your fixed point is above (since your x^2 term is positive) the parabola. In this case it will be impossible to draw a line that is parallel to the parabola.
 
  • #9
epenguin
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Hmm, you mean the two lines will have different slopes? If so, then that is exactly what I was expecting. Can you elaborate on how sketching this helps me?

As a rule for math problems, don't hold a debate with yourself before doing a sketch! :oldbiggrin:
 

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