Slope of the tangent to the curve `x=at^(2), y=2t` at `t=2` is ……………….. .

Answer

To find the slope of the tangent to the curve defined by the parametric equations \(x = at^2\) and \(y = 2t\) at \(t = 2\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)
We need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

1. Differentiate \(x = at^2\):
\[ \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at \]

2. Differentiate \(y = 2t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(2t) = 2 \]

### Step 2: Find \(\frac{dy}{dx}\)
Using the chain rule, we can express \(\frac{dy}{dx}\) in terms of \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2at} = \frac{1}{at} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(t = 2\)
Now we substitute \(t = 2\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}\bigg|_{t=2} = \frac{1}{a(2)} = \frac{1}{2a} \]

### Conclusion
The slope of the tangent to the curve at \(t = 2\) is:
\[ \frac{1}{2a} \]

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