Question

The locus the centre of the circle described on any chord of a parabola ${y^2} = 4ax$ as diameter is

$$a){x^2} = 2a(4 - a) \\\ b){x^2} = - 2a(y - a) \\\ c){y^2} = 2a(x - a) \\\ d){y^2} = - 2a(x - a) \;$$

Answer 1

Hint : Here the given question is of coordinate geometry including circle and parabola, to solve this question you have to know the properties of parabola which defines the value of extremities of the chord drawn.

Complete step-by-step answer :
The given question is to find the locus of the circle described on any chord of a parabola as its diameter, here we have to first assume the extremities of parabola then after assuming the centre of the circle and solving these equations we can reach to the final solution.
Hence let extremities of the focal chord of the given parabola ${y^2} = 4ax$ is :
$\Rightarrow A = (a{t_1}^2,a{t_2}^2)\,and\,B = (a{t_1}^2,2a{t_2})$
Here we know that
$\Rightarrow {t_1} \times {t_2} = - 1$
Now we have to assume the centre of the circle, let say (c,d) be the coordinate of the centre.
Here we know that is the circle is described on the focal cord then AB will be the diameter of the circle, and then centre of the circle can be written as:
$\Rightarrow d = \dfrac{a}{2}({t_1}^2 + {t_2}^2)\,and\,c = a({t_1} + {t_2})$
Now to eliminate these variable of “t” we have square both side of the coordinates value for “c” and then on putting values from coordinate “d” we can reach to the final solution:

$$\Rightarrow {c^2} = {a^2}({t_1}^2 + {t_2}^2 + 2{t_1}{t_2}) \\\ \Rightarrow {c^2} = {a^2}\left( {\dfrac{{2d}}{a} - 2} \right)\\{ \sin ce\,{t_1}^2 + {t_2}^2 = \dfrac{{2d}}{a}from\,circle\,coordinates\\} \\\ \Rightarrow {c^2} = 2a(d - a) \;$$

Now here for general equation of circle:
We have to replace the assume coordinate of centre with (x,y):

$$\Rightarrow c \to y,d \to x \\\ \Rightarrow {y^2} = 2a(x - a) \;$$

This is our final required equation for the circle.
So, the correct answer is “${y^2} = 2a(x - a)$”.

Note : Here the given question needs to be solved by using properties of parabola which gives the coordinates of the extremities of any cord in the parabola, and after that we can simply solve for the equation formed and get the solution.