As it is known that the equation of the parabola can be properly presented by the quadratic equation. The quadratic equation can be described as the function of x, therefore, the same can be represented it by f(x). Now, parabola touches the x axis at (1,0). Therefore, 1 is considered to be the root of f(x). So, f (1) =0. Now, on the other hand, if f (1) is considered to be zero, therefore, it cannot be able to pass through the point of (1,1) and it would mean that f(1)=1. Now, it is seen that 1 cannot be made equal to zero and therefore, the parabola do not exist:
F: (n, k)
√ (x-n)2 + (y-k)2 = y-mx
(x-n)2 + (y-k)2 = (y-mx)2
X= 0 and y = 1- (1)
h2 + (k-1)2 = 1/1+m2
X= 1 and y=0
(h-1)2 + k2 = m2/1+m2 – (2)
Adding both the equations 1 and 2,
H2 + k2 – 2k + 1+ h2 – 2h + 1 + k2 = 1+m2/1+m2= 1
2h2 + 2k2-2h-2k+1=0
Therefore, A movable parabola touches x-axis and y-axis at (0,1) and (1,0). Then the locus of the focus of the parabola is 2x2+2y2-2x-2y+1=0.
You will receive a confirmation email shortly in your subscribe email address.
You have already subscribed our newsletter.