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A road follows the curve y = x² - 2x + 3, with x and y in kilometres…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A road follows the curve $y = x^2 - 2x + 3$ , with $x$ and $y$ in kilometres. A straight railway cuts the road at the points $(1, 2)$ and $(3, 6)$ . Find the point of the road where its direction is exactly the same as the direction of the railway.

Need a reminder? The key ideaShow

A tangent parallel to a line of gradient $m$ has $f'(x) = m$ . A perpendicular tangent has $f'(x) = -\dfrac{1}{m}$ . For a tangent from an outside point, call the point of contact $(t, f(t))$ . The angle between curves satisfies $\tan\varphi = \left|\dfrac{m_1 - m_2}{1 + m_1m_2}\right|$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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