Answer: \(y=(x-2)^2+4\); \(\frac{dy}{dx}=2(x-2)\); tangent \(y-8=4(x-4)\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

(a) Since

\(x=2+\operatorname{sec}\theta,\)

we have

\(\operatorname{sec}\theta=x-2.\)

Using

\(\tan^2\theta=\operatorname{sec}^2\theta-1,\)

the expression for \(y\) becomes

\(y=5+\operatorname{sec}^2\theta-1.\)

So

\(y=4+(x-2)^2.\)

(b) Differentiate:

\(\frac{dy}{dx}=2(x-2).\)

(c) When \(\theta=\frac{\pi}{3}\),

\(\operatorname{sec}\frac{\pi}{3}=2.\)

So

\(x=2+2=4.\)

Then

\(\tan^2\frac{\pi}{3}=3,\)

so

\(y=5+3=8.\)

The gradient at \(x=4\) is

\(\frac{dy}{dx}=2(4-2)=4.\)

Therefore the tangent is

\(y-8=4(x-4).\)