The normal at a point on the parabola y^2=4x | vedclass

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Question

(B) The equation of the parabola is $y^2=4ax$,where $a=1$. The equation of a normal to the parabola $y^2=4ax$ at point $t$ is $y = -tx + 2at + at^3$.

Vedclass · Accepted Answer

$(2,0)$

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(B) The equation of the parabola is $y^2=4ax$,where $a=1$. The equation of a normal to the parabola $y^2=4ax$ at point $t$ is $y = -tx + 2at + at^3$. Since the normal passes through $(5,0)$,we have $0 = -5t + 2(1)t + (1)t^3$. $t^3 - 3t = 0 \implies t(t^2 - 3) = 0$. The values of $t$ are $t_1 = 0, t_2 = \sqrt{3}, t_3 = -\sqrt{3}$. The feet of the normals are $(at^2, 2at)$,which are: $P_1 = (1(0)^2, 2(1)(0)) = (0,0)$. $P_2 = (1(\sqrt{3})^2, 2(1)(\sqrt{3})) = (3, 2\sqrt{3})$. $P_3 = (1(-\sqrt{3})^2, 2(1)(-\sqrt{3})) = (3, -2\sqrt{3})$. The centroid $C$ of the triangle formed by these points is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$. $C = \left(\frac{0+3+3}{3}, \frac{0+2\sqrt{3}-2\sqrt{3}}{3}\right) = \left(\frac{6}{3}, 0\right) = (2,0)$.

Process	Condition
$(I)$ Adiabatic	$(A) \Delta W = 0$
$(II)$ Isothermal	$(B) \Delta Q = 0$
$(III)$ Isochoric	$(C) \Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D) \Delta U = 0$

The normal at a point on the parabola $y^2=4x$ passes through $(5,0)$. If two more normals to this parabola also pass through $(5,0)$,then the centroid of the triangle formed by the feet of these three normals is

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