A curve has parametric equations \(x=5a\sec\theta,\,y=3a\tan\theta\), where \(-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi\) and \(a\) is a positive constant. Find the coordinates of the point on the curve at which the normal is parallel to the line \(y=x\).

Differentiating with respect to \(\theta\), we find that \[\begin{align*} \frac{dx}{d\theta} &= 5a\sec\theta\tan\theta \\ \frac{dy}{d\theta} &= 3a\sec^2\theta . \end{align*}\]

Thus \[\frac{dy}{dx}=\frac{3a\sec^2\theta}{5a\sec\theta\tan\theta}=\frac{3}{5\sin\theta}.\]

The normal to the curve therefore has gradient \[-\frac{5}{3}\sin\theta.\]

The normal is parallel to \(y=x\) when \[-\frac{5}{3}\sin\theta=1\quad \Longrightarrow \quad \sin\theta=-\frac{3}{5}.\]

By sketching a graph of \(y=\sin\theta\), we can see that there’s a unique value of \(\theta\) in the range \(-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}\) for which \(\sin\theta=-\dfrac{3}{5}\).

The blue line in the diagram shows that this corresponds to a positive value of \(\cos\theta\).

Hence, we get that \[\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} \quad \Longrightarrow \quad \cos \theta = \frac{4}{5}\] as \(\cos\theta\) is positive.

A right-angled triangle with angle labelled theta, opposite side of length 3 and hypotenuse of length 5, with the adjacent side of length 4 (using the Pythagorean theorem).
\(y=\sin\theta\) (red), \(y=\cos\theta\) (black) and the value of \(\theta\) for which the normal is parallel to \(y=x\) (blue)

Alternatively, using the sketch above, we can see that we are dealing with a \(3-4-5\) triangle, and so the value of cosine we need is \(\cos\theta=\dfrac{4}{5}\).

Thus our coordinates for this value of \(\theta\) are \[\begin{align*} x &= 5a\sec\theta = \frac{5a}{\cos\theta} = \frac{5a\times 5}{4} = \frac{25a}{4}, \\ y &= 3a\tan\theta=3a \times \frac{\sin \theta}{\cos \theta} = 3a\times\left(-\frac{3}{5}\right)\times\left(\frac{5}{4}\right) = -\frac{9a}{4}. \end{align*}\]