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Tangent and Curves

  1. Jan 12, 2007 #1
    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.
    Last edited: Jan 13, 2007
  2. jcsd
  3. Jan 12, 2007 #2


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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.
  4. Jan 13, 2007 #3
    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?
  5. Jan 13, 2007 #4


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    Okay, that looks good now. You final equation will have a parameter t in it since this is "for any t".
  6. Jan 14, 2007 #5

    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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