# Tangent and Curves

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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HallsofIvy
Homework Helper
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?

HallsofIvy