Finding Perpendicular Tangent Point on Parabola

In summary, To find the point on the parabola y= 4x^2 + 2x - 5 where the tangent line is perpendicular to the line 3x + 2y = 7, you need to find the slope of the given line and use the definition of a derivative to find the x value where the derivative equals the negative reciprocal of the slope. This will give you the x value of the point, which you can then plug into the original equation to find the y value. In this case, the point is (-1/6, 13/3).
  • #1
soul5
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Homework Statement


Find the point on the parabola y= 4x^2 + 2x - 5 where the tangent line is perpendicular to the line 3x + 2y = 7.


Homework Equations





The Attempt at a Solution


I don't know what to do since I was away the last 3 classes since I was away. Help me please.
 
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  • #2
You want to find the slope of the line you're given and use the definition of a derivative as well as perpendicularity to solve for the x value you want.
 
  • #3
Two perpendicular lines have slopes that are negative reciprocals of each other, eg: a line with a slope 2 is perpendicular to a line with a slope -1/2.

Find the slope of the line, find the negative reciprocal of that slope. The derivative of a function is the slope of that graph at any point on the graph, so find the derivative of the parabola and see at what value of x it will equal the negative reciprocal of the slope you found earlier.
 
  • #4
TMM said:
You want to find the slope of the line you're given and use the definition of a derivative as well as perpendicularity to solve for the x value you want.

So I take the slope of this? 3x + 2y = 7

so...

2y = -3x + 7
y= -3/2x + 7/2

slope = -3/2 so if it is perpendicular the slope is 2/3 is that my right slope?

I now I have to do more but it that right so far?
 
  • #5
Yes.
 
  • #6
Correct.

dy/dx = 8x+2
You want the value of x when dy/dx is (2/3), as you said from above.
Solving for x gets (-1/6).
Plug this value into your original equation y=4x^2 etc.
 

1. What is a line tangent to a parabola?

A line tangent to a parabola is a straight line that touches the curve of the parabola at exactly one point. This point of contact is known as the point of tangency.

2. How is the slope of a line tangent to a parabola calculated?

The slope of a line tangent to a parabola can be calculated using the derivative of the parabola's equation. The derivative is a mathematical function that represents the rate of change of the parabola at a specific point.

3. Can a line be tangent to a parabola at more than one point?

No, a line can only be tangent to a parabola at one point. This is because a parabola is a symmetrical curve and has a unique slope at each point on the curve.

4. How does the position of the line tangent to a parabola change with respect to the parabola's vertex?

The position of the line tangent to a parabola changes as the parabola's vertex moves. When the vertex is at the origin, the tangent line is parallel to the x-axis. As the vertex moves to the right or left, the tangent line also moves accordingly.

5. What is the significance of a line tangent to a parabola in real-life applications?

In real-life applications, a line tangent to a parabola is used to find the maximum or minimum value of a function. This is because the tangent line is perpendicular to the parabola's curve at the point of tangency, and the slope of the tangent line represents the rate of change of the function at that point.

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