<s>[INST] Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992) 
(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)
 [/INST] 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.
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