Mathematics: 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.

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Consider the parabola r(t)equalsleft angle at squared

goodkida7

12.03.2022

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>

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