Answer: \(y=10\mathrm e x-4\mathrm e\), and \(x_P=\dfrac25\).

Differentiate the curve:

\(\frac{dy}{dx}=10\mathrm e^{2x-1}.\)

At \(x=1\), the gradient is

\(\left.\frac{dy}{dx}\right|_{x=1}=10\mathrm e.\)

The corresponding \(y\)-coordinate is

\(y=5\mathrm e^{2(1)-1}+\mathrm e=6\mathrm e.\)

So the tangent passes through \((1,6\mathrm e)\) with gradient \(10\mathrm e\).

Its equation is

\(y-6\mathrm e=10\mathrm e(x-1).\)

Therefore

\(y=10\mathrm e x-4\mathrm e.\)

At the \(x\)-axis, \(y=0\), so

\(0=10\mathrm e x-4\mathrm e.\)

Thus \(10x=4\), and

\(x_P=\frac25.\)

This completes the solution and gives the required result.