Question

Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity​, where a is a positive real number. Find all points on the parabola at which r and bold r prime are orthogonal.

Answer 1

Given:-   $r(t)=< at^2+1,t>$  ; $-\infty < t< \infty$ , where a is any positive real number.

Consider the helix parabolic equation :  

                                              $r(t)=< at^2+1,t>$

now, take the derivatives we get;

                                            $r{}'(t)=$

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  $r(t) and r{}'(t)$  are orthogonal i.e,   $r\cdot r{}'=0$

Therefore, we have ,

                                  $< at^2+1,t>\cdot < 2at,1>=0$

$< at^2+1,t>\cdot < 2at,1>=$

                                              $=2a^2t^3+2at+t$

$2a^2t^3+2at+t=0$

take t common in above equation we get,

$t\cdot \left (2a^2t^2+2a+1\right )=0$

⇒$t=0$ or $2a^2t^2+2a+1=0$

To find the solution for t;

take $2a^2t^2+2a+1=0$

The number$D = b^2 -4ac$ determined from the coefficients of the equation $ax^2 + bx + c = 0.$

The determinant $D=0-4(2a^2)(2a+1)$$=-8a^2\cdot(2a+1)$

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  $r(t)=< at^2+1,t>$ i.e <1,0>