Solving Curve 8y^2=x^3 & Finding Tangent & Point Q

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Homework Help Overview

The discussion revolves around the curve defined by the equation 8y² = x³ and the exploration of a point P parameterized by x = 2t² and y = t³. Participants are tasked with finding the tangent line at point P and determining where this tangent intersects the curve again at point Q.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to show that point P lies on the curve for all values of t and seeks to find the tangent line at that point. Some participants question the validity of the point being on the curve and suggest verifying the equations by substituting the parameterization into the curve's equation.

Discussion Status

Participants are actively discussing the correctness of the initial parameterization and the subsequent calculations for the tangent line. There is acknowledgment of a mistake in the original problem statement, and some guidance has been offered regarding the approach to finding the tangent and the intersection point Q.

Contextual Notes

There is a noted confusion regarding the parameterization of point P and its relationship to the curve, with participants exploring whether the problem was copied correctly. The presence of a parameter t in the final equations is also a point of discussion.

Harmony
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Show that for all values of t, the point P with the equation
x=2t^2, y=t^3 lies on the curve 8y^2=x^3

Find the equation of the tangent to the curve at point P.

The tangent meets the curve once again at point Q. Find the coordinates of point Q.

I can find the equation of the tangent.

After finding the tangent, I try to solve it by using the two equation. At the end I get 9t^2x^2-12t^4x+4t^6=2x^3. Is it the right way? It seems complicated.
 
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I'm moving this to "caculus and beyond"- finding tangents in general is basic to calculus.

How did you get that equation? You certainly shouldn't have both t and x in the equation: to show that the given point is on the curve for all t, replace x in the equation by 2t3 and replace y by t3. Is the result an equation that is true for all t?

Actually it isn't. 8y2 is 8t6 while x3 is 8t9: the point is NOT on the curve for all t. In fact, if x= 2t3 and y= t3 then, obviously, x= 2y. that's just the straight line y= x/2. Are you sure you have copied the problem correctly? For example, if x= 2t2, then it would be true. Or it might be that you are asked to find a value of t such that P is on the curve: the point where the straight line y= x/2 crosses the curve 8y2= x3.

I suspect you are asked to find the value of t (find P) such that P is on the curve because asking you to find the "tangent to the curve at point P" doesn't make much sense if P is any point on the curve.
 
I am terribly sorry...there was a mistake when I typed the question. x=2t^2.
To obtain the tangent I just power the x by 3 and power the y by 2 and cancel off the t. There wasn't much trouble in that.
Equation of the tangent : 4y=3tx-2t^3
Since the two equation intersect, I approach this question by substituting y from the equation 8y^2=x^3. That's how I get 9t^2x^2-12t^4x+4t^6=2x^3, and find that it is hard for me to obtain the coordinate of Q through such equation. Am I on the right track?
 
Okay, that looks good now. You final equation will have a parameter t in it since this is "for any t".
 
9t^2x^2-12t^4x+4t^6=2x^3

But I can't solve this equation...even with the parameter t. I suspect there is something wrong with my calculation.
 

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