The sum of diameters of the circles that touch (i) the parabola $$75x^2 = 64(5y - 3)$$ at the point $$\left(\dfrac{8}{5}, \dfrac{6}{5}\right)$$ and (ii) the $$y$$-axis, is equal to ______.

Given parabola,

$$75x^2=64(5y-3)$$

Point of contact is

$$P\left(\frac85,\frac65\right)$$

Differentiate implicitly:

$$150x=320\frac{dy}{dx}$$

$$\frac{dy}{dx}=\frac{15x}{32}$$

At

$$x=\frac85,$$

slope of tangent is

$$m=\frac{15(8/5)}{32}=\frac34$$

Hence tangent at $$P$$ is

$$y-\frac65=\frac34\left(x-\frac85\right)$$

Simplifying,

$$3x-4y=0$$

Let

$$L=3x-4y$$

Now family of circles touching the parabola at $$P$$ is

$$\left(x-\frac85\right)^2+\left(y-\frac65\right)^2+\lambda(3x-4y)=0$$

Expanding,

$$x^2+y^2-\frac{16}{5}x-\frac{12}{5}y+4+3\lambda x-4\lambda y=0$$

$$x^2+y^2+\left(3\lambda-\frac{16}{5}\right)x-\left(4\lambda+\frac{12}{5}\right)y+4=0$$

Comparing with

$$x^2+y^2+2gx+2fy+c=0,$$

we get

$$2g=3\lambda-\frac{16}{5}$$

$$2f=-\left(4\lambda+\frac{12}{5}\right)$$

and

$$c=4$$

Since the circle touches the $$y$$-axis,

$$\text{radius}=|g|$$

Using

$$r^2=g^2+f^2-c,$$

we get

$$g^2+f^2-c=g^2$$

$$f^2=c$$

Thus,

$$\left(2\lambda+\frac65\right)^2=4$$

Hence,

$$2\lambda+\frac65=\pm2$$

Case 1:

$$2\lambda+\frac65=2$$

$$2\lambda=\frac45$$

$$\lambda=\frac25$$

Diameter is

$$|2g| =\left|3\left(\frac25\right)-\frac{16}{5}\right| =|-2| =2$$

Case 2:

$$2\lambda+\frac65=-2$$

$$2\lambda=-\frac{16}{5}$$

$$\lambda=-\frac85$$

Diameter is

$$|2g| =\left|3\left(-\frac85\right)-\frac{16}{5}\right| =|-8| =8$$

Therefore, sum of diameters is

$$2+8=10$$

Hence, the required answer is

$$\boxed{10}$$