Elliptic Curve Cryptography insecure when input does not lie on the curve? I am new to Elliptic Curve Cryptography and I was reading up on it online when I came across this link. It stated the following.


  Unfortunately, there is a gap between ECDLP difficulty and ECC security. None of these standards do a good job of ensuring ECC security. There are many attacks that break real-world ECC without solving ECDLP. The core problem is that if you implement the standard curves, chances are you're doing it wrong:
  
  
  Your implementation produces incorrect results for some rare curve points.
  Your implementation leaks secret data when the input isn't a curve point.
  Your implementation leaks secret data through branch timing.
  Your implementation leaks secret data through cache timing.
  


So, I was curious about the second point. How is it possible to leak secret data when the input isn't a curve point. I am assuming it means the base point(P) is not on the curve, but then how would P+P be calculated and how would it make the implementation insecure?
Suppose you publish a public key $P = [n]G$ for a secret scalar $n$, where $G$ is the standard base point.

If you are willing to tell me $H([n]Q)$ given $Q = (x, y)$ for any coordinates $x$ and $y$ of my choice, then I can send you a point $Q$ of (say) order 2 on some other curve whose arithmetic law happens to coincide with the curve you meant to use, and learn $H([n]Q) = H([n \bmod 2]Q)$.

How does this work?  Suppose you use the short Weierstrass form $y^2 = x^3 + a_4 x + a_6$.  Since the addition law doesn't involve the constants $a_4$ or $a_6$ at all, and the doubling law involve only $a_4$, it turns out that you will happily compute scalar multiplication of points on any curve with the same $a_4$, so all I need to do is find some $a'_6$ whose curve has low-order points.

When I get back the answer, it is either $H([0]Q) = H(\mathcal O)$ or $H([1]Q) = H(Q)$, so by examining which one you returned, I can learn $n \bmod 2$.  The same applies if I send a point of order 3, 4, 5, etc., and I can combine the results with the Chinese remainder theorem to interactively compute the discrete log $n$ of your public key $P = [n]G$ with your help.

Of course, if you expose the point $[n]Q$ instead of the hash $H([n]Q)$, then my task may be even easier!