Review question

# Can we show that these normals intersect on this circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9413

## Solution

Prove that the equation of the normal at a point of the curve $x=n\cos t-\cos nt,\qquad y=n\sin t-\sin nt,$ where $n$ is an integer greater than unity, is $x\cos \frac{n+1}{2}t + y\sin \frac{n+1}{2}t=(n-1)\cos \frac{n-1}{2}t.$

First we need the gradient of the curve and from that the gradient of the normal. We use the chain rule to differentiate the parametric equation.

We have that $x=n\cos t-\cos nt,$ so $\frac{dx}{dt}=-n\sin t+n\sin nt$ and $y=n\sin t-\sin nt,$ so $\frac{dy}{dt}=n\cos t-n\cos nt.$ Now we could find the gradient of the curve since $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

The form for the answer we are trying to prove is quite different, so it’s worth having a think before ploughing on.

We notice that the form of the answer contains arguments of $\frac{n+1}{2}t$ or $\frac{nt+t}{2}$ which points us toward the following identities

\begin{align} \sin a-\sin b&= 2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\label{trig1} \\ \cos a-\cos b&= 2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{b-a}{2}\right) \label{trig2} \\ \cos a \cos b&= \frac{1}{2}\cos(a-b)+\frac{1}{2}\cos(a+b) \label{trig3} \\ \sin a \sin b&= \frac{1}{2}\cos (a-b) -\frac{1}{2}\cos(a+b) \label{trig4} \end{align}

Alternatively, we could use the more familiar compound angle identities. Then some slightly different algebra would lead to the same result.

Applying identity $\eqref{trig1}$, we have that $\frac{dx}{dt}=2n\sin\left(\frac{n-1}{2}t\right)\cos\left(\frac{n+1}{2}t\right).$

Applying identity $\eqref{trig2}$ yields $\frac{dy}{dt}=2n\sin\left(\frac{n+1}{2}t\right)\sin\left(\frac{n-1}{2}t\right).$

and so $\frac{dy}{dx}=\frac{2n\sin\left(\frac{n+1}{2}t\right)\sin\left(\frac{n-1}{2}t\right)}{2n\sin\left(\frac{n-1}{2}t\right)\cos\left(\frac{n+1}{2}t\right)}=\frac{\sin \left(\frac{n+1}{2}t\right)}{\cos \left(\frac{n+1}{2}t\right)}.$ Therefore the gradient of the normal to the curve is $$$-\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)} \label{eq:grad-of-norm}$$$ and the normal to the curve has equation $y-\left( n\sin t -\sin nt \right)=-\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)}\bigl(x-\left(n\cos t-\cos nt\right)\bigr).$ Multiplying by $\sin\left(\frac{n+1}{2}t\right)$ gives $\sin\left(\frac{n+1}{2}t\right)y-\sin\left(\frac{n+1}{2}t\right)\left( n\sin t -\sin nt \right)=\quad$ $\quad-\cos\left(\frac{n+1}{2}t\right)x+\cos\left(\frac{n+1}{2}t\right)\left(n\cos t-\cos nt\right)$ and rearranging gives us \begin{align*} \sin\left(\frac{n+1}{2}t\right)y+\cos\left(\frac{n+1}{2}t\right)x &= \sin\left(\frac{n+1}{2}t\right)\left( n\sin t -\sin nt \right) \\ &+\cos\left(\frac{n+1}{2}t\right)\left(n\cos t-\cos nt\right). \end{align*} Multiplying out the brackets and applying identities $\eqref{trig3}$ and $\eqref{trig4}$ we find that \begin{align*} \sin\left(\frac{n+1}{2}t\right)y+\cos\left(\frac{n+1}{2}t\right)x&=n\cos t\cos\left(\frac{n+1}{2}t\right)-\cos nt\cos\left(\frac{n+1}{2}t\right)\\ &\quad+n\sin t\sin\left(\frac{n+1}{2}t\right)-\sin nt\sin\left(\frac{n+1}{2}t\right) \\ &=\frac{n}{2}\left[\cos\left(\frac{n-1}{2}t\right)+\cos\left(\frac{n+3}{2}t\right)\right] \\ &\quad -\frac{1}{2}\left[\cos\left(\frac{n-1}{2}t\right)+\cos\left(\frac{3n+1}{2}t\right)\right] \\ &\quad +\frac{n}{2}\left[\cos\left(\frac{n-1}{2}t\right)-\cos\left(\frac{n+3}{2}t\right)\right] \\ &\quad -\frac{1}{2}\left[\cos\left(\frac{n-1}{2}t\right)-\cos\left(\frac{3n+1}{2}t\right)\right] \\ &=(n-1)\cos\left(\frac{n-1}{2}t\right), \end{align*}

which is exactly what we wanted.

Show that, if $n$ is an even integer, the normals at the points $t$ and $t+\pi$ are perpendicular…

From $\eqref{eq:grad-of-norm}$ we have that the gradient of the normal to the curve at $t$ is $m_N(t) = -\frac{\cos \left(\frac{n+1}{2}t\right)}{\sin \left(\frac{n+1}{2}t\right)}.$

Let’s use a shorthand $A=\dfrac{n+1}{2}t$ and $P=\dfrac{n+1}{2}\pi$. Then we can write the gradients of the normals at $t$ and $t+\pi$ as $m_N(t) = -\frac{\cos A}{\sin A} \quad\text{and}\quad m_N(t+\pi) = -\frac{\cos (A+P)}{\sin (A+P)}.$

Multiplying these two gradients together (we hope to get $-1$) and then using identities $\eqref{trig3}$ and $\eqref{trig4}$, we have \begin{align*} m_N(t) \times m_N(t+\pi) &= \frac{\cos A \cos (A+P)}{\sin A \sin (A+P)} \\ &= \frac{\cos(-P)+\cos(2A+P)}{\cos(-P)-\cos(2A+P)} \end{align*}

Since $n$ is an even integer, $P=\frac{\pi}{2}, \frac{3\pi}{2}, \cdots$ so $\cos(-P) = 0$. The fraction therefore cancels down, the product of the two gradients is indeed $-1$ and the normals are perpendicular, as required.

…and intersect on the circle $x^2+y^2=(n-1)^2.$

Using our shorthand $A$ and $P$ from above and also defining $B=\dfrac{n-1}{2}t$, we can write the equations of the two normals as \begin{align} x\cos A + y\sin A &= (n-1)\cos B \label{eq:n_t}\\ \text{and}\quad x\cos (A+P) + y\sin (A+P) &= (n-1)\cos (B+P-\pi) \label{eq:n_t+pi} \end{align}

The form of the equation of the circle we are aiming for implies that we want to square these equations, but first let’s expand the $(A+P)$ expressions.

Using the compound angle identities we have \begin{align*} \cos(A+P) &= \cos A \cos P - \sin A \sin P = - \sin A \sin P \\ \sin(A+P) &= \sin A \cos P + \cos A \sin P = \cos A \sin P \\ \cos(B+P-\pi) &= \cos B \cos (P-\pi) - \sin B \sin(P-\pi) = \sin B \sin P \end{align*}

We have used the facts that $\cos P = \cos(P-\pi)=0$ and $\sin(P-\pi) = -\sin P$. Notice also that $\sin P$ is either $1$ or $-1$ depending on the value of $n$.

Equation $\eqref{eq:n_t+pi}$ then becomes \begin{align*} -x\sin A \sin P + y\cos A \sin P &= (n-1)\sin B \sin P \\ \implies\quad -x\sin A + y\cos A &= (n-1)\sin B \end{align*} Squaring this equation and equation $\eqref{eq:n_t}$ then adding the results we get \begin{align*} x^2\sin^2 A + y^2\cos^2A-2xy\sin A\cos A &= (n-1)^2\sin^2 B \\ \text{and}\quad x^2\cos^2 A + y^2\sin^2A+2xy\sin A\cos A &= (n-1)^2\cos^2 B \\ \implies\quad x^2\left(\sin^2A+\cos^2A\right) + y^2\left(\cos^2A+\sin^2A\right) &= (n-1)^2\left(\sin^2B+\cos^2B\right) \\ \implies\quad x^2 + y^2 &= (n-1)^2 \end{align*}

as required.