Finding the slope of line tangent to a parabola

  • Thread starter oates151
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  • #1
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Homework Statement



Find the equations of both lines through the point (2,-3) that are tangent to the parabola y=(x^2)+x

Homework Equations





The Attempt at a Solution



Took the derivative and got a slope of 5 and the slope of the normal line being -1/5, but the answer was marked wrong. How do I do this?

Two equations I got
y=-(1/5)x-(13/5)
y=5x-13
 

Answers and Replies

  • #2
ElijahRockers
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Choose a point (a,y(a)). A line that goes through this point, AND the given point must have a slope of [itex]\frac{y(a) - (-3)}{a - 2}[/itex] Also, the slope at point 'a' can be given by the derivative of the function. This gives you two equal expressions for the slope in terms of a. It will be a quadratic equation. The roots will be the x values at which the lines intersect the parabola.
 
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  • #3
SammyS
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What do you mean? "both lines"

One of your lines is tangent to the parabola at (2, -3) .

The other is normal to the parabola at (2, -3) .
 
  • #4
ElijahRockers
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What do you mean? "both lines"

One of your lines is tangent to the parabola at (2, -3) .

The other is normal to the parabola at (2, -3) .

Actually I think it's tangent to the parabola at (2, 6)
 
  • #5
SammyS
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Actually I think it's tangent to the parabola at (2, 6)
Ha!

Yup, the parabola doesn't pass through (2, -3) ! DUH
 

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