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.

Given:-    ; , where a is any positive real number.

Consider the helix parabolic equation :  

now, take the derivatives we get;

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

Here,    are orthogonal i.e,   

Therefore, we have ,

take t common in above equation we get,

⇒ or

To find the solution for t;

take

The number determined from the coefficients of the equation

The determinant

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>

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step-by-step explanation:

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)