Given:-
;
, where a is any positive real number.
Consider the helix parabolic equation :
![r(t)=< at^2+1,t>](https://tex.z-dn.net/?f=r%28t%29%3D%3C%20at%5E2%2B1%2Ct%3E)
now, take the derivatives we get;
![r{}'(t)=](https://tex.z-dn.net/?f=r%7B%7D%27%28t%29%3D%3C2at%2C1%3E)
As, we know that two vectors are orthogonal if their dot product is zero.
Here,
are orthogonal i.e, ![r\cdot r{}'=0](https://tex.z-dn.net/?f=r%5Ccdot%20r%7B%7D%27%3D0)
Therefore, we have ,
![< at^2+1,t>\cdot < 2at,1>=0](https://tex.z-dn.net/?f=%3C%20at%5E2%2B1%2Ct%3E%5Ccdot%20%3C%202at%2C1%3E%3D0)
![< at^2+1,t>\cdot < 2at,1>=](https://tex.z-dn.net/?f=%3C%20at%5E2%2B1%2Ct%3E%5Ccdot%20%3C%202at%2C1%3E%3D%3Cat%5E2%2B1%5Ccdot%5Cleft%282at%5Cright%29%2C%20t%5Ccdot%20%5Cleft%281%29%3E)
![=2a^2t^3+2at+t](https://tex.z-dn.net/?f=%3D2a%5E2t%5E3%2B2at%2Bt)
![2a^2t^3+2at+t=0](https://tex.z-dn.net/?f=2a%5E2t%5E3%2B2at%2Bt%3D0)
take t common in above equation we get,
![t\cdot \left (2a^2t^2+2a+1\right )=0](https://tex.z-dn.net/?f=t%5Ccdot%20%5Cleft%20%282a%5E2t%5E2%2B2a%2B1%5Cright%20%29%3D0)
⇒
or ![2a^2t^2+2a+1=0](https://tex.z-dn.net/?f=2a%5E2t%5E2%2B2a%2B1%3D0)
To find the solution for t;
take ![2a^2t^2+2a+1=0](https://tex.z-dn.net/?f=2a%5E2t%5E2%2B2a%2B1%3D0)
The number
determined from the coefficients of the equation ![ax^2 + bx + c = 0.](https://tex.z-dn.net/?f=ax%5E2%20%2B%20bx%20%2B%20c%20%3D%200.)
The determinant ![D=0-4(2a^2)(2a+1)](https://tex.z-dn.net/?f=D%3D0-4%282a%5E2%29%282a%2B1%29)
![=-8a^2\cdot(2a+1)](https://tex.z-dn.net/?f=%3D-8a%5E2%5Ccdot%282a%2B1%29)
Since, for any positive value of a determinant is negative.
Therefore, there is no solution.
The only solution, we have t=0.
Hence, we have only one points on the parabola
i.e <1,0>