\(y-{p}^{2}x = \frac{5}{p}-5{p}^{3}\) crosses the line \(y = -x\) at point A
Show that point A is \(\left ( \frac{5}{p}-5{p}^{3},-\frac{5}{p}+5{p}^{3} \right )\)
This was the last question in my further maths paper, couldn't do it
+33653 Replace y by -x
-x - p^2*x = 5/p - 5p^3
-x*(1 + p^2) = (5/p)*(1 - p^4)
-x*(1 + p^2) = (5/p)*(1 + p^2)*(1 - p^2)
-x = (5/p)*(1 - p^2)
x = -5/p + 5p
Hence y = 5/p - 5p
Hmm, doesn't match your desired solution!
+129840 I got the same result as Alan did.....as a check......
y - p^2x ....... let x = 5p - 5/p ..... and let y = -x .... and we have....
[5/p - 5p] - p^2 [ 5p - 5/p] =
5/p - 5p - 5p^3 + 5p =
5/p - 5p^3 and this = the right side
So.......the solution should be :
(x , y) = ( 5p - 5/p , 5/p - 5p )
