The discussion centers on the mathematical problem of finding the normal line at point P(ap^2, 2ap) on the parabola defined by the equation y^2 = 4ax and determining its intersection with the curve at point Q(aq^2, 2aq). Participants concluded that the equation p^2 + pq + 2 = 0 must be shown, alongside the locus of intersection of tangents at points P and Q, which is expressed as y^2(x + 2a) + 4a^3 = 0. Several users encountered difficulties in simplifying expressions and verifying the correctness of the problem statement.
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The expression is also dividable by q-p.sooyong94 said:Homework Statement
If the normal at P(ap^2 ,2ap) to the parabola y^2 = 4ax meets the curve again at Q(aq^2, 2aq), show that p^2 +pq+2=0
Homework Equations
Point-slope form
The Attempt at a Solution
I tried putting y=2aq and x=aq^2 but I can seem to simplify the whole thing other than dividing both sides by a
There should be minus in front of the last term in the first equation. You made a mistake when copying.sooyong94 said:Strangely enough I got this:
You have to work with x and y. What are p and q now?sooyong94 said:
Have you copied the question correctly? These x and y values do not fulfill the equation given y^2(x+2a)+4a^3 =0.sooyong94 said:
I mean the problem might be wrong. The coordinates of the point of intersection do not fit to the given locus.sooyong94 said: