Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992)

Question

(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin B )$ should subtend a right angle at O is $$a^2\cos A\cos B+b^2\sin A\sin B=0$$ (ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at the point $(C\cos t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ subtend a right angle at $O$.

Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$

Part (ii) gives the result $x\cos t+y\sin t=C$

With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.

I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.

Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)

Answer 1

The work is indeed tedious.

Suppose that the tangent cuts the ellipse at $P(a\cos A, b\sin A)$ and $Q(a\cos B, b\sin B)$, then the equation of the chord $PQ$ is:

$$\frac{y-b\sin B}{x-a\cos B}=\frac{b(\sin B-\sin A)}{a(\cos B-\cos A)}$$ \

$$\iff ay(\cos B-\cos A) - ab\sin B(\cos B-\cos A)= bx(\sin B-\sin A)-ab(\sin B-\sin A)$$

\ $$\iff bx(\sin B-\sin A)+ay(\cos A-\cos B)=ab\sin (B-A)$$ \

$$\iff bx\left(2\sin \frac{B-A}{2}\cos \frac{A+B}{2}\right)+ay\left(2\sin \frac{B-A}{2}\sin \frac{A+B}{2}\right)=ab\left(2\sin \frac{B-A}{2}\cos \frac{B-A}{2}\right)$$ \

$$\iff bx\left(\cos \frac{A+B}{2}\right)+ay\left(\sin \frac{A+B}{2}\right)=ab\left(\cos \frac{B-A}{2}\right)$$ \ $$\iff bx\left(\frac {\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}\right)+ay\left(\frac {\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}\right)=\frac{ab}{\sqrt{a^2+b^2}}$$

Since the tangent cuts the ellipse at $P$, $Q$

This is the same as the equation of the tangent

$$x \cos t + y \sin t = \frac{ab}{\sqrt{a^2+b^2}}$$

Comparing the two equations, we have

$$\cos t = \frac {b\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}$$

and

$$\sin t = \frac {a\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}}$$

Thus we have $$\left(\frac {b\cos \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}} \right)^2+\left(\frac {a\sin \frac{A+B}{2}}{\sqrt{a^2+b^2}\cdot \cos \frac{B-A}{2}} \right)^2=1$$ \ $$b^2\cos^2\left(\frac{A+B}{2} \right)+a^2\sin^2\left(\frac{A+B}{2} \right)=\left(a^2+b^2 \right)\cos^2 \left(\frac {B-A}{2}\right)$$ \ $$a^2\left[\cos^2\left(\frac {B-A}{2}\right)-\sin^2\left(\frac{A+B}{2} \right)\right]+b^2\left[\cos^2\left(\frac {B-A}{2}\right)-\cos^2\left(\frac{A+B}{2} \right)\right]=0$$ \ $$a^2\left[\frac{1+\cos(B-A)}{2}-\frac{1-\cos(A+B)}{2} \right]+b^2\left[\frac{1+\cos(B-A)}{2}-\frac{1+\cos(A+B)}{2}\right]=0$$ \ $$a^2\left[\cos(B-A)+\cos(A+B)\right] + b^2\left[\cos(B-A)-\cos(A+B) \right]=0$$ \ $$a^2(2\cos A \cos B)+ b^2(2\sin A \sin B)=0$$ \ $$a^2\cos A \cos B+ b^2\sin A \sin B=0$$

which is the condition for $OP$ and $OQ$ to be perpendicular.

Answer 2

Your idea of parametrising the ellipse works. We write a general point of the ellipse in the form $(a\cos\psi,b\sin\psi)$ for some angle $\psi$. Since the problem is invariant under scaling, we set $a^2+b^2=1$ without loss of generality, so $C=ab$. (This is just for convenience, the solution works just fine without this.) A generic point on the tangent line can be written as $$ (C\cos t-\lambda \sin t,C\sin t+\lambda\cos t) $$ for some real parameter $\lambda$; since there are two points of intersections let them correspond to $\lambda_1,\lambda_2$. We have the equations $$ \begin{cases}a\cos\psi = C\cos t -\lambda\sin t\\ b\sin\psi = C\sin t+\lambda\cos t.\end{cases} $$ Now, taking ratios, we find that $$ \frac{b}{a}\tan\psi = \frac{C\sin t+\lambda\cos t}{C\cos t-\lambda\sin t} = \frac{\tan t+\lambda/C}{1-(\lambda/C)\tan t} = \tan(t+\phi) $$ where angle $\phi$ is defined so that $\tan\phi=\lambda/C$. By the given (i) of the question, we just need to show that the two values $\psi_1,\psi_2$ corresponding to the two points satisfy $$ a^2\cos\psi_1\cos\psi_2+b^2\sin\psi_1\sin\psi_2=0 \iff 1+\frac{b^2}{a^2}\tan\psi_1\tan\psi_2=0. $$ Using the relation for $\psi$ we have found, we rewrite this as $$ \left(\frac{b}{a}\tan\psi_1\right)\left(\frac{b}{a}\tan\psi_1\right)=-1 \iff \tan(t+\phi_1)\tan(t+\phi_2)=-1. $$ By standard properties of the tangent function, this is easily seen to be equivalent to the condition that $\phi_1=\phi_2-\pi/2$ (up to multiples of $\pi$). Taking $\tan$ of both sides, we see that it is equivalent to $x_1=-1/x_2$, where $x_i := \tan(\phi_i) = \lambda_i/C$ ($i=1,2$), i.e. $x_1x_2=-1$.

Now, we will endeavour to show this. By the equation of the ellipse, we have $$ b^2(C^2\cos^2t-2C\lambda\cos t\sin t+\lambda^2\sin^2 t)+a^2(C^2\sin^2t+2C\lambda\cos t\sin t+ \lambda^2\cos^2 t)=a^2b^2. $$ Substituting $C=ab$ and $a^2+b^2=1$, simplifying, then collecting terms, with $x=\lambda/C$, we end up with the quadratic $$ (b^2\sin^2 t+a^2\cos^2t)x^2 + 2(a^2-b^2)(\sin t\cos t)x + b^2\cos^2t+a^2\sin^2t-1=0. $$ Now, we finally find that the product of roots of this equation $$ \begin{split}x_1x_2 &= \frac{b^2\cos^2t+a^2\sin^2t-1}{b^2\sin^2 t+a^2\cos^2t}\\ &= \frac{(1-a^2)(1-\sin^2t)+a^2\sin^2t-1}{(1-a^2)\sin^2t+a^2(1-\sin^2t)} \\ &= \frac{1-a^2-\sin^2t+2a^2\sin^2t-1}{\sin^2t+a^2-2a^2\sin^2t} = -1,\end{split} $$ as desired, so we are done.

Answer 3

Another way: I might say this way is "more obvious" in the sense that it is easy to think of. You "just" sit down and crunch through the algebra as if it were an ordinary simultaneous equation job. Though, one must be very careful with the algebra.

The cases where $\cos(t)=0$ or $\sin(t)=0$ can be very easily handled. Suppose now that neither is zero. Then we arrange the tangent line as: $$y=C\csc(t)-x\cot(t)$$And the ellipse equation can be arranged as: $$b^2x^2+a^2y^2-a^2b^2=0$$Let $y=C\csc(t)-x\cot(t)$ in the ellipse equation: $$\begin{align}b^2x^2+a^2[C^2\csc^2(t)+x^2\cos^2(t)\csc^2(t)-2C\cos(t)\csc^2(t)\cdot x]-a^2b^2&=0\\x^2[b^2\sin^2(t)+a^2\cos^2(t)]+a^2C^2-a^2b^2\sin^2(t)-2a^2C\cos(t)\cdot x&=0\\\frac{x^2}{a^2}[b^2\sin^2(t)+a^2\cos^2(t)]+\frac{a^2b^2}{a^2+b^2}-b^2\sin^2(t)-\frac{2ab\cos(t)}{\sqrt{a^2+b^2}}\cdot x&=0\end{align}$$Now let $x=a\cos\vartheta$ for some $\vartheta$ to be determined: $$\cos^2\vartheta[b^2\sin^2(t)+a^2\cos^2(t)]+b^2\cdot\frac{a^2\cos^2(t)-b^2\sin^2(t)}{a^2+b^2}-\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\cos\vartheta=0$$We use the quadratic formula. The "$b^2-4ac$" term involves a very nice difference-of-two-squares: $$\begin{align}\frac{4b^2}{a^2+b^2}a^4\cos^2(t)-&\frac{4b^2}{a^2+b^2}[b^2\sin^2(t)+a^2\cos^2(t)][a^2\cos^2(t)-b^2\sin^2(t)]\\&=\frac{4b^2}{a^2+b^2}[a^4\cos^2(t)(1-\cos^2(t))+b^4\sin^4(t)]\\&=\frac{4b^2}{a^2+b^2}\sin^2(t)\cdot[a^4\cos^2(t)+b^4\sin^2(t)]\end{align}$$The quadratic formula then yields: $$\begin{align}\cos\vartheta&=\frac{\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\pm\frac{2b\sin(t)}{\sqrt{a^2+b^2}}\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{2[b^2\sin^2(t)+a^2\cos^2(t)]}\\&=\frac{b}{\sqrt{a^2+b^2}}\cdot\frac{a^2\cos(t)\pm\sin(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^2\sin^2(t)+a^2\cos^2(t)}\end{align}$$A near-identical computation, arising from letting $x=C\sec(t)-y\tan(t)$, cancelling the $\sec^2$ terms, and dividing by $b^2$ to get a quadratic formula in $\sin\vartheta$, then solving, gives: $$\sin\vartheta=\frac{a}{\sqrt{a^2+b^2}}\cdot\frac{b^2\sin(t)\pm\cos(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^4\sin^2(t)+a^4\cos^2(t)}$$Now we use the hint of part $(i)$ and the difference of two squares: $$\begin{align}a^2\cos(\vartheta_+)\cos(\vartheta_-)+b^2\sin(\vartheta_+)\sin(\vartheta_-)&=\underset{K}{\underbrace{\left(\frac{ab}{[b^4\sin^2(t)+a^4\cos^2(t)]\sqrt{a^2+b^2}}\right)^2}}\\&\times\Big[a^4\cos^2(t)-\sin^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\\&+b^4\sin^2(t)-\cos^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\Big]\\&=K\cdot\Big[a^4\cos^2(t)(1-\sin^2(t))+b^4\sin^2(t)(1-\cos^2(t))\\&-a^4\cos^4(t)-b^4\sin^4(t)\Big]\\&=0\end{align}$$

So the intersection points involving $a\cos\vartheta_{\pm},b\sin\vartheta_{\pm}$ (the pairing of which will vary in general) subtend a right angle, as desired.

Answer 4

(i) We have $P = (a \cos A, b \sin A)$ and $Q = (a \cos B, b \sin B) $

Since $OP \perp OQ$ , then by using the dot product we get

$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $

(ii) A point on the circle with radius $C$ centered at the origin is parameterized as

$ P = C (\cos t, \sin t)$

The tangent vector at $P$ is along the vector $(-\sin t , \cos t ) $ (Perpendicular to the radius vector).

Hence, the parametric equation of the tangent is

$ \ell(t,s) = (C \cos t - s \sin t , C \sin t + s \cos t ) $

(iii) Now the above tangent line is to intersect the ellipse, and we expect two intersection points. Let $P = (a \cos A, b \sin A) $ be the intersection point, then $P$ lies on $\ell$.

Hence

$ a \cos A = C \cos t - s \sin t \hspace{30pt} b \sin A = C \sin t + s \cos t $

Solving for $s$ from the above two equations

$ s = \dfrac{ a \cos A - C \cos t }{-\sin t} = \dfrac{ b \sin A - C \sin t }{\cos t } $

Cross-multiplying yields

$ a \cos A \cos t - C \cos^2 t = - b \sin A \sin t + C \sin^2 t $

So that

$ a \cos A \cos t + b \sin A \sin t = C \hspace{30pt}(1) $

Similarly, if $Q= (a \cos B, b \sin B)$ is the other intersection point, then

$ a \cos B \cos t + b \sin B \sin t = C \hspace{30pt}(2)$

Equations $(1)$ and $(2)$ are linear in $\cos t $ and $\sin t $. Therefore, we can solve for them, and using Cramer's rule (for example), or any other method, we get

$ \cos t = \dfrac{ C b (\sin B - \sin A) }{ ab (\cos A \sin B - \cos B \sin A)} $

$ \sin t = \dfrac{ C a (\cos A - \cos B) }{ ab (\cos A \sin B - \cos B \sin A) } $

Now we can use the fact that $\cos^2 t + \sin^2 t = 1 $ to write

$ C^2 \bigg(b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 \bigg) = (ab)^2 \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{15pt}(3)$

Now, we're given that $ C = \dfrac{ab}{\sqrt{a^2 + b^2} } $

Substituting this into $(3)$,

$b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 = (a^2 + b^2) \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{20pt}(4)$

Expanding both the left and right sides of $(4)$,

$ b^2 (\sin^2 B + \sin^2 A - 2 \sin A \sin B) + a^2 (\cos^2 A + \cos^2 B - 2 \cos A \cos B) = (a^2 + b^2) (\cos^2 A \sin^2 B + \cos^2 B \sin^2 A - 2 \cos A \cos B \sin A \sin B ) $

$ b^2 \sin^2 B (1 - \cos^2 A) + b^2 \sin^2 A (1 - \cos^2 B) + a^2 \cos^2 A (1 - \sin^2 B) + a^2 \cos^2 B (1 - \sin^2 A) - 2 b^2 \sin A \sin B - 2 a^2 \cos A \cos B + 2 (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $

And this becomes, after using $ \cos^2 A + \sin^2 A = 1 $ and $ \cos^2 B + \sin^2 B = 1 $, and dividing through by $2$:

$ b^2 \sin^2 A \sin^2 B + a^2 \cos^2 A \cos^2 B - a^2 \cos A \cos B - b^2 \sin A \sin B + (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $

Collecting like terms,

$ (a^2 \cos A \cos B + b^2 \sin A \sin B) ( \sin A \sin B + \cos A \cos B - 1) = 0 $

But $\sin A \sin B + \cos A \cos B - 1 = \cos(A - B) - 1 \ne 0 $ (because $A \ne B$).

Therefore,

$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $

And from part (i), this shows that the intersection points subtend a right angle at the origin.