Finding a Tangent Point with a Given Slope on a Cubic Curve

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In summary, to find a point P on the curve y=x3 where the tangent line at P intersects the curve again at point Q with a slope that is 4 times the slope at P, we can use the equations y'=3x2 and the slope formula (y2-y1)/(x2-x1). By setting 2P=Q and using algebra to express Q in terms of P, we can find two unknowns a and c. We can then write the equation of the tangent line at P and use the given slope condition to solve for the values of a and c.
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stonecoldgen
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Homework Statement


Find a point P on y=x3 such that the tangent at P intersects the original curve again at point Q so that the slope of the tangent at Q is 4 times the slope of the tangent at P.


Homework Equations


y'=3x2

and the slope of the what I think it should be a secant line=(y2-y1)/(x2-x1)

The Attempt at a Solution



I figured that 2P=Q because 3P2=(3/4)Q2 and then by using algebra I expressed Q in terms of P.

I know this leads to somewhere, but I am not sure what should I do next.
 
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But P and Q are points. So P2 and Q2 don't make any sense. Try writing P = (a,a3) and Q= (c,c3) so you have two unknowns a and c. Write the equation of the tangent line at P and use the two facts that it must intersect Q and the slope condition you are given.
 

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