Can we find the equation of the normal to the curve when $t = 2$?

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We can write \[\begin{equation*} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \end{equation*}\] We have that (by the Chain Rule) \[\begin{equation*} \frac{dx}{dt} = 2(3-2t)(-2) = 4(2t-3), \quad \frac{dy}{dt} = 2t - 2 \end{equation*}\] and therefore \[\begin{equation*} \frac{dy}{dx} = \frac{2t - 2}{4(2t-3)} = \frac{t-1}{2(2t-3)}. \end{equation*}\]

The stationary point of the curve occurs when \(\dfrac{dy}{dx} = 0\), i.e. when \(t = 1\).

As \(x = (3-2t)^2\) and \(y = t^2 - 2t\), the co-ordinates of the stationary point are \((1,-1)\).

When \(t = 2\), the gradient of the curve is \[\begin{equation*} \frac{dy}{dx} = \frac{2-1}{2(2 \times 2 - 3)} = \frac{1}{2}. \end{equation*}\]

The gradient of the normal to the curve at this point is the negative of the reciprocal of this, i.e. \(-2\).

Thus, the equation of the normal is \(y = -2x + c\), for some constant \(c\), where we require that the original curve crosses the normal at \(t=2\).

When \(t=2\), the curve has \(x\)-coordinate \((3 - 2 \times 2)^2 = 1\) and \(y\)-coordinate \(2^2 - 2 \times 2 = 0\).

So, substituting these values into the equation for the normal, we require \(c\) to be such that \(0 = - 2 \times 1 + c\), i.e. \(c = 2\).

Consequently, the equation of the normal is \(y = -2x + 2\).