2004 pure maths P2 ellipse,急!

你這個問題很有意思。因為我沒有marking或者standard solution在手﹐所以我自己先整條題目做了一次。結果我自己(c)(i)都是用了y = mx + c 去證。既然三個人都用slope﹐看來slope是沒有問題的。不過我看了你的方法也是沒有問題的﹐你是先做了(c)(ii)再證到(c)(i)。不過我又不覺得用slope複雜﹐因為一般由圓外一點求切線這個都是常用方法﹐只是煩罷了﹐但是是routine的。這是其中一個原因﹐第二個原因是跟回問題的思路。當然我不認為你的方法會被扣分。

(a) Since

(acosθ)^2/a^2 + (bsinθ)^2/b^2 = 1

We shows that P lies on E

(b)(i) The tangent of E at P(acosθ, bsinθ) is

xcosθ/a + ysinθ/b = 1....(1)

(ii) Sub. x = 0 and y = 0 into (1) respectively. We get

P1(a/cosθ, 0) ; P2(0, b/sinθ)

(c)(i) Let the equation of L1 is y = m1x + c1

Then by Sub. P1, we have c1 = -am1/cosθ

Sub. L1 into C1

x^2 + (m1x + c1)^2 = a^2

(1 + m1^2)x^2 + 2m1cx + (c^2 - a^2) = 0

Determinant = 0

4m1^2c1^2 - 4(1 + m1^2)(c1^2 - a^2) = 0

m1^2c1^2 - (c1^2 - a^2 + m1^2c1^2 - m1^2a^2) = 0

m1^2a^2 + a^2 - c1^2 = 0

m1^2a^2 + a^2 - [a^2m1^2/(cosθ)^2] = 0

m1^2 = (cosθ)^2/[1 - (cosθ)^2] = (cotθ)^2

m1 = -cotθ (because m1 < 0 and 0< θ < π/2, cotθ > 0)

Similarly, let the equation of L1 is y = m2x + c2

We can find that c2 = b/sinθ

From m2^2b^2 + b^2 - c2^2 = 0

m2^2 + 1 - 1/(sinθ)^2 = 0

m2^2 = (cotθ)^2

m2 = -cotθ

So, L1 is parallel to L2

(ii) Let Q1(acosθ1, asinθ1)

Then m1 = -x/y | (acosθ1, asinθ1) = -cotθ1

Since m1 = -cotθ and both θ and θ1 <π/2, we have θ1 = θ

So, Q1(acosθ, asinθ)

Similarly, Q2(bcosθ, bsinθ)

(iii) The slope of l is (bsinθ - asinθ)/(bcosθ - acosθ) = tanθ and we can see that l is perpendicular to both L1 and L2. So, l is common normal to C1 and C2.

我認為這樣做法完全沒有問題.