Luke M.

asked • 10/29/22

How is the equation of the secant line parallel to the tangent line found?

Find the equation of the secant line Parallel to the tangent line to the curve −5x8+x28y2+y8=−1    passing the point (1,1).

Luke M.

−5x^8+x^28y^2+y^8=−1
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10/29/22

Doug C.

Is that x^(28) or x^2 (8y^2)
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10/29/22

Doug C.

Actually I graphed them both on Desmos and neither passes through (1,1), so double check your post.
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10/29/22

Luke M.

Hi Doug, here is the exact question I'm trying to answer: Find the equation of the secant line Parallel to the tangent line to the curve f(x) = -5x^8+x^(28)*y^2+y^8=-1 passing the point (1,1)
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10/29/22

Doug C.

OK, I think I get it. (1,1) is not the point of tangency it is a point NOT on the curve, but a point which a tangent line to the curve passes through.
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10/29/22

Luke M.

I know my slope from d/dx of the function to be 6/5 through the point (1,1), I'm just confused by the question, I've spoken to 3 tutors at my university center and 2 professors, one in which is the writer of the assignment who informed me with this in an email: Hi Luke, Your answer is correct , it was a typo mistake and i fixed it , the question is: Find the equation of the secant line Parallel to the tangent line to the curve passing through (1,1). You would have a slope of the line (through the curve) (6/5) and a point (1,1). Thank you, Prof. Mulvey
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10/29/22

Doug C.

This does not make any sense to me. You implicitly find dy/dx and get something like [(2x^7)(10-7x^20y^2)]/[x^28y+4y^7]. That is a formula for slopes of tangent lines to the graph of the original relation. I agree when you substitute x=1 and y =1 into that formula for slope, you do get 6/5. But for that to be the slope of the tangent line passing through (1,1), the point (1,1) must be on the original curve. But if you plug in (1,1) you get an answer of -3 (not -1), so (1,1) is not on the curve.
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10/29/22

Doug C.

As a matter of fact for the given relation when x = 1, the y value must be approximately 1.13304089329 in order to get a relation value close to -1. Feels like this is a misguided problem. To get the slope of a tangent line to the original relation, let (a,b) be a point on the curve that has a tangent line passing through the point (1,1). Then the slope of the line through (1,1) and (a,b) is equal to (b-1)/(a-1). And when you substitute (a,b) into the formula for dy/dx those expressions for slope must be equal. That gives you one equation with two variable. The other equation comes from substituting (a,b) into the original relation. Now you have two equations and two unknowns with exponents like 28 and 27. Finding the point of tangency would involve solving that system for a and b. Once you have those values, you can plug into dy/dx to find the slope of the tangent line tangent to the curve AND passing through (1,1). Now all secant lines parallel to that tangent line will have the same slope. Seems to me there will be infinitely many of those. Or is it the secant line that passes through (1,1)? The problem as suggested by Prof Mulvey needs much more clarification (IMHO).
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10/29/22

Luke M.

Here is the link to the video that was posted by the professor to answer the question, if it doesn't help you understand my assignment, I'm at a complete loss in direction https://www.youtube.com/watch?time_continue=24&v=3J0m53qkb0k&feature=emb_logo
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10/29/22

Dayv O.

how dy/dx = 6/5 is a mystery to me. dy/dx is the tangent slope at some given point on the curve. No point is given. But it is simple algebra to say secant line with slope 6/5 through (1,1) is (6/5)=(y-1)/(x-1) as long as curve's curvature is toward (1,1). Secant line must be inside of curve and intersect curve at two points.
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10/29/22

Doug C.

Here is a graph that shows a secant line that passes through the point (1,1) and is parallel to a point on the curve where the x coordinate is also 1. Note that in this case the slope is NOT 6/5. desmos.com/calculator/qsc1xvjywp
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10/29/22

Luke M.

I appreciate the help in this confusingly worded exercise
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10/29/22

David M.

I don't believe the exercise is "confusingly" worded. I believe the exercise was either written with the wrong coefficients in the equation or with the wrong point for the tangent line to pass through. Another issue I have is that since parallel lines do not intersect, the secant cannot be the same line as the tangent and cannot pass through the same point; and without a specific point assigned to the secant, there must be an infinite number of secant lines passing through two points on the curve being, at the same time, parallel to (same slope/derivative as) the tangent line, unless Doug C.'s graphical analysis of the equation shows a uniqueness that allows for a possible singular secant line for some tangent line to the curve (which I am not taking the time to check out).
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11/07/22

1 Expert Answer

By:

Jonathan T. answered • 10/26/23

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To find the equation of the secant line parallel to the tangent line to the curve at a given point, follow these steps:


1. Find the derivative of the curve equation to determine the slope of the tangent line.


2. Calculate the slope of the tangent line at the given point.


3. Use the point-slope form of a line to find the equation of the tangent line.


4. Determine the same slope for the secant line to keep it parallel to the tangent line.


5. Use the point-slope form again to find the equation of the secant line.


Let's apply these steps to the curve equation -5x^8 + x^28y^2 + y^8 = -1 and the point (1, 1).


1. Find the derivative of the curve equation to determine the slope of the tangent line:


Differentiate the curve equation with respect to x:


-40x^7 + 56x^27yy' + 8y^7y' = 0


2. Calculate the slope of the tangent line at the point (1, 1):


Plug in the coordinates (x, y) = (1, 1) into the derivative:


-40(1)^7 + 56(1)^27(1)y' + 8(1)^7y' = 0

-40 + 56y' + 8y' = 0


Combine like terms:


64y' - 40 = 0


Solve for y':


64y' = 40

y' = 40/64

y' = 5/8


So, the slope of the tangent line at (1, 1) is 5/8.


3. Use the point-slope form of a line to find the equation of the tangent line:


y - y1 = m(x - x1), where (x1, y1) is the point on the curve.


y - 1 = (5/8)(x - 1)


Now, we have the equation of the tangent line:


y = (5/8)(x - 1) + 1


4. Determine the same slope (5/8) for the secant line to keep it parallel to the tangent line.


5. Use the point-slope form again to find the equation of the secant line passing through (1, 1):


y - 1 = (5/8)(x - 1)


This is the equation of the secant line parallel to the tangent line at the point (1, 1) on the curve -5x^8 + x^28y^2 + y^8 = -1.

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Doug C.

Except for the fact that (1,1) is not on the curve, hence the 13 comments to this question when it was first posed. The student worked to get clarification from the professor, but it did not really happen.
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10/26/23

Jonathan T.

There should be enough information for the student to figure out questions going forward. Best of luck!
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10/26/23

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