Find the equation of a tangent line to y = x^2?

  • #1
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Homework Statement


the line goes through (0, 3/2) and is orthogonal to a tangent line to the part of parabola y = x^2, x > 0

Homework Equations




The Attempt at a Solution


I have problems regarding finding the equation of tangent line to the part of parabola
because the question not specifically mention at which point

we know that the tangent line (y2) tangent to the part of parabola
so, the slope of y2 = the slope of parabola at (x,y)

parabola y = x^2
so the slope is 2x
and the slope changes according to x

which x should I used for x > 0?
if I use x = 1, y = 1^2= 1, slope = 2
and the y2 would be
y - 1 = 2(x - 1)
y = 2x - 1

but if I used x = 3, y = 9, slope = 6
and the y2 would be
y - 9 = 6(x-3)
y = 6x - 9

and y2 is different for each x...

I need to find the slope of y2, so I can find the line equation orthogonal to y2 passing through (0, 3/2) ...
 

Answers and Replies

  • #2
scottdave
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Do you know the relationship between the slopes of perpendicular (orthogonal) lines?
This article will help. http://www.purplemath.com/modules/slope3.htm

If you know the relationship between slopes of orthogonal lines, then you can get an expression for the slope of a line orthogonal to the parabola. They give you what point to use.
 
  • #3
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Do you know the relationship between the slopes of perpendicular (orthogonal) lines?
This article will help. http://www.purplemath.com/modules/slope3.htm

If you know the relationship between slopes of orthogonal lines, then you can get an expression for the slope of a line orthogonal to the parabola. They give you what point to use.

for y1 orthogonal to y2
means
slope of y1 * slope of y2 = -1

how to find the equation of a line orthogonal to the parabola?
y - y1 = m(x - x1)
what's the slope? and which point?

slope of parabola = 2x
 
  • #4
CWatters
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A tangent to a curve has the same slope as the curve at the point of contact. Do you know how to find an equation for the slope of a curve?
 
  • #5
scottdave
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What expression would you use for m of the orthogonal line, if the parabola has a slope of 2x ? How can we take this information and create an equation for a line (yes m will seem to be variable). Now find the intersection point between that line, and the parabola.
 
  • #6
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A tangent to a curve has the same slope as the curve at the point of contact. Do you know how to find an equation for the slope of a curve?
slope of the curve is 2x

What expression would you use for m of the orthogonal line, if the parabola has a slope of 2x ? How can we take this information and create an equation for a line (yes m will seem to be variable). Now find the intersection point between that line, and the parabola.
m for parabola = 2x
m for the tangent line is 2x or 2?
the m for orthogonal line is -1/(2x) or -1/2
 
  • #7
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Thread moved. Questions involving derivatives belong in the Calculus & Beyond section, not the Precalc section.
 
  • #8
Ray Vickson
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Do you know the relationship between the slopes of perpendicular (orthogonal) lines?
This article will help. http://www.purplemath.com/modules/slope3.htm

If you know the relationship between slopes of orthogonal lines, then you can get an expression for the slope of a line orthogonal to the parabola. They give you what point to use.

The OP is correct: he needs to know which tangent to use, because given any tangent you can find a unique line orthogonal to that tangent and passing through the point (0,3/2). However, if the intersection of the line and the tangent actually lies on the curve (in other words, if the line to be drawn passes both through (0,3/2) and the tangency point) then, indeed, there is a unique solution.
 
  • #9
scottdave
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This is the way that I approached it. Let x = a on the parabola y = x2, so the point (a,a2) is on the parabola. The tangent line at that point has a slope of 2a, and the orthogonal line has a slope of -1/(2a). Draw a line from this point to the specified point (0,3/2) and solve for a.
 
  • #10
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The OP is correct: he needs to know which tangent to use, because given any tangent you can find a unique line orthogonal to that tangent and passing through the point (0,3/2). However, if the intersection of the line and the tangent actually lies on the curve (in other words, if the line to be drawn passes both through (0,3/2) and the tangency point) then, indeed, there is a unique solution.
This is the way that I approached it. Let x = a on the parabola y = x2, so the point (a,a2) is on the parabola. The tangent line at that point has a slope of 2a, and the orthogonal line has a slope of -1/(2a). Draw a line from this point to the specified point (0,3/2) and solve for a.

thanks, i think i get your point
 
Last edited:
  • #11
SammyS
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Homework Statement


the line goes through (0, 3/2) and is orthogonal to a tangent line to the part of parabola y = x^2, x > 0
Please give the complete statement of the problem as it was given to you.
 
  • #12
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Please give the complete statement of the problem as it was given to you.
Consider a parabola y = x^2. Answer the following questions and fill in your responses in the corresponding boxes on the answer sheets

(1) the line that goes through the point (0, 3/2) and is orthogonal to a tangent line to the part of parabola y = x^2 with x > 0 is y = Ax + 3/2
Solve for A
 

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