# Quod Erat Demonstrandum

## 2009/03/26

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 12:09 上午

A F.5 student asked me the following question some days ago, reply now.

A(-3,0) and B(-1,0) are two points and P(x,y) is a variable point such that $PA = \sqrt{3}PB$. Let C be the locus of P.

(a) Show that the equation of C is $x^2 + y^2 = 3$.

(b) T(a,b) is a point on C. Find the equation of the tangent to C at T.

(c) The tangent from A to C touches C at a point S in the second quadrant. Find the coordinates of S.

(d) L is a straight line which passes through point A and makes an angle $\theta$ with the positive $x$-axis, where $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. Q(x,y) is a point on L such that $AQ = r$. (See the figure below)

(i) Write down the coordinates of Q in terms of r and $\theta$.

(ii) L cuts C at two distinct points H and K. Let $AH = r_1$, $AK = r_2$.

(1) Show that $r_1, r_2$ are roots of the quadratic equation $r^2 - 6r\cos\theta + 6 = 0$.

(2) Find the range of possible values of $\theta$, giving your answers correct to three significant figures.

(HKCEE 1999)

(a) By the fact that $PA = \sqrt{3}PB$, it is done trivially.

(b) Immediately, $ax + by = 3$ is the answer.

(c) Let $y = m(x + 3)$ be a tangent to C, then put it into the $x^2 + y^2 = 3$, obtain a quadratic equation in $x$. For the tangency, set $\Delta = 0$, we will have the coordinates of the point of contact are ($-1,\sqrt{2}$) as the answer because the point is in the second quadrant.

(d) This is about the parametric form of coordinates of Q. That is, we use one parameter $r$ to describe the position of Q. Different values of $r$, corresponding to different positions of Q.

(i) Let R be the foot of Q on the x-axis, then $AR = r\cos\theta$, hence the x-coordinate of Q = $-3 + r\cos\theta$; also, $QR = r\sin\theta$; therefore the y-coordinate of Q is $r\sin\theta$.
Thus Q($-3 + r\cos\theta$ , $r\sin\theta$).

(ii) (1) Students may have difficulties in doing this part. In most of the cases, to find the points of intersection of a line and a circle (say), they are trained to substitute the equation of the line into the equation of circle. But in this question, the equation of line is not so ‘explicit’, it is given by so-called parametric form. That is,

$Q(-3 + r\cos\theta , r\sin\theta)$ represents an equation of a straight line actually. For different values of $r$, we have different position of $Q$, and the locus of $Q$ is actually a straight line, explicitly, the $x$ and $y$ coordinates of the line are given as

$x = -3 + r\cos\theta$
$y = r\sin\theta$

To find the points of intersection of the line and the circle, we just put the above equations into $x^2 + y^2 = 3$, yield

$(-3 + r\cos\theta)^2 + (r\sin\theta)^2 = 3$
$9 - 6r\cos\theta + r^2\cos^2\theta + r\sin^2\theta - 3 = 0$
$r^2 - 6r\cos\theta + 6 = 0$ – – – – – – (*)

which is a quadratic equation in $r$ (not $\theta$, $\theta$ is fixed).

So what is the (geometric) meaning of the roots of (*)?

Urm, suppose (*) has two distinct roots (two $r$-values) which are corresponding to two distinct points of intersections between the line L and the circle C (or two distinct positions of the moving point Q).

Now, it is given that, L really cuts C at two distinct points H and K, such that, $AH = r_1$ and $AK = r_2$, right? That is, the corresponding $r$-values (precisely, parameters) of H and K are $r_1$ and $r_2$ respectively. Thus, $r_1$ and $r_2$ are roots of (*).

(d)(ii)(2) For L cuts C at two distinct points, there should be two distinct roots in (*), hence $\Delta > 0$. Thus

$(-6\cos\theta)^2 - 4(6) > 0$
$\cos^2\theta > \frac{2}{3}$
$-\cos^{-1}(\sqrt{\frac{2}{3}}) < \theta < \cos^{-1}(\sqrt{\frac{2}{3}})$

It may not be easy to yield the last inequality above, just sketch a graph to see it.

F.5 students, don’t turn a blind eye to this problem, it IS IN THE SYLLABUS.

## 1 則迴響 »

1. it is very nice and it is help us to learn more and more

迴響 由 mehezabeen — 2009/03/31 @ 2:04 上午 | 回應