What will be the value of the co-ordinate whose position of a particle moving along the parabola y2 = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?

The correct option is (4, 4).

Step-by-Step Explanation:

Let the position of the particle at any time $t$ be represented by the coordinates $(x,y)$, where $x$ is the abscissa and $y$ is the ordinate.

The particle moves along the parabola given by the equation:
$$y^2=4x$$
Let us label this as Equation (1).

We are given that the rate of increase of the abscissa ($x$) with respect to time ($t$) is twice the rate of increase of the ordinate ($y$) with respect to time ($t$). Mathematically, this relation is:
$$\frac{dx}{dt}=2\frac{dy}{dt}$$
Let us label this as Equation (2).

To find the coordinates of the point, we differentiate Equation (1) with respect to time $t$ using the chain rule:
$$\frac{d}{dt}\left(y^2\right)=\frac{d}{dt}\left(4x\right)$$
$$2y\frac{dy}{dt}=4\frac{dx}{dt}$$
Let us label this as Equation (3).

Now, substitute the value of $\frac{dx}{dt}$ from Equation (2) into Equation (3):
$$2y\frac{dy}{dt}=4\left(2\frac{dy}{dt}\right)$$
$$2y\frac{dy}{dt}=8\frac{dy}{dt}$$
Assuming the rate of change of the ordinate is non-zero ($\frac{dy}{dt}\ne 0$), we can divide both sides by $\frac{dy}{dt}$:
$$2y=8$$
Dividing by 2 gives:
$$y=4$$

Now, we find the corresponding $x$-coordinate by substituting $y=4$ back into Equation (1):
$$4^2=4x$$
$$16=4x$$
Dividing both sides by 4:
$$x=4$$

Thus, the position of the particle at this instant is $(4,4)$.