The equation of  tangent to the curve \(x=\dfrac{2at^2}{1+t^2}, y=\dfrac{2at^3}{1+t^2}\) at the point for which \(t=\dfrac{1}{2}\), is

Concept:

Equation of a tangent to the curve is given by (y - y₁) = m(x - x₁); where m is the slope of the curve i.e., \(\frac{dy}{dx}=m\).

Calculation:

Given 

\(x=\dfrac{2at^2}{1+t^2}, y=\dfrac{2at^3}{1+t^2}\)

As these x and y are in terms of parameter t

In order to find the slope i.e.,\(m=\frac{dy}{dx}\) , we shall differentiate x with respect to t and y with respect to t separately.

So, first, let us find \(\frac{dx}{dt}\) by differentiating x with respect to t by using the quotient rule

\(\frac{dx}{dt}=\frac{d}{dx}\big(\frac{2at^2}{1+t^2}\big)\) .....(1)

⇒ \(\frac{dx}{dt}=\frac{(1+t^2).4at-(2at^2).2t}{(1+t^2)^2}\)

⇒ \(\frac{dx}{dt}=\frac{4at}{(1+t^2)^2}\) 

Now, first, let us find \(\frac{dy}{dt}\) by differentiating x with respect to t by using the quotient rule

\(\frac{dy}{dt}=\frac{d}{dx}\big(\frac{2at^3}{1+t^2}\big)\)

⇒ \(\frac{dy}{dt}=\frac{(1+t^2)(6at^2)-(t^3).2t}{(1+t^2)^2}\)

⇒ \(\frac{dy}{dt}=\frac{6at^2+2at^4}{(1+t^2)^2}\)   ......(2)

By using (1) and (2) we have,

\(\frac{dy}{dx}=\frac{6at^2+2at^4}{4at}\)

⇒ \(\frac{dy}{dx}=\frac{3t+t^3}{2}\)

At t = 1/2 we  have

\(\frac{dy}{dx}=\frac{13}{16}\)

Also at t = 1/2, we get 

\(x =\frac{2a}{5}=x_1, y=\frac{a}{5}=y_1\)

The equation of a tangent to a curve is given by (y - y1) = m(x - x1)

(or) (y - y1) = \(\frac{dy}{dx} \)(x - x1)

By using all the values and substituting them in the above equation we get,

\(y-\frac{a}{5}=\frac{13}{16}(x-\frac{2a}{5})\)

on simplifying this we get,

13x - 16y = 2a.

Hence, the correct answer is option 1)