Problems

Filters
Clear Filters

1 problem found

2005 Paper 1 Q6
D: 1500.0 B: 1490.2
coordinate geometry circle distance formula locus Apollonius circle simultaneous equations ratio of distances

  1. The point \(A\) has coordinates \(\l 5 \, , 16 \r\) and the point \(B\) has coordinates \(\l -4 \, , 4 \r\). The variable point \(P\) has coordinates \(\l x \, , y \r\,\) and moves on a path such that \(AP=2BP\). Show that the Cartesian equation of the path of \(P\) is \[ \displaystyle \l x+7 \r^2 + y^2 =100 \;. \]
  2. The point \(C\) has coordinates \(\l a \, , 0 \r\) and the point \(D\) has coordinates \(\l b \, , 0 \r\), where \(a\ne b\). The variable point \(Q\) moves on a path such that \[ QC = k \times QD\;, \] where \(k>1\,\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[ \frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}\;. \] Show further that \((a+7)(b+7)=100\,\).

Solution:

  1. Since \(AP = 2BP\) we also have \(|AP|^2 = 4|BP|^2\) ie \begin{align*} && (x-5)^2 + (y-16)^2 &= 4(x+4)^2 + 4(y-4)^2 \\ \Rightarrow && x^2 - 10x+25 + y^2 -32y + 256 &= 4x^2+32x+64+4y^2-32y+64 \\ \Rightarrow && 281 &= 3x^2+42x+3y^2+128\\ && 281 &= 3(x+7)^2-147+3y^2+128 \\ \Rightarrow && 300 &= 3(x+7)^2 + 3y^2 \\ && 100 &= (x+7)^2 + y^2 \end{align*}
  2. Since \(|QC|^2 = k^2 |QD|^2\), \begin{align*} && (x-a)^2 + y^2 &= k^2 (x-b)^2 + k^2y^2 \\ \Rightarrow && x^2-2ax+a^2 &= k^2x^2-2k^2bx+k^2b^2 + (k^2-1)y^2 \\ && a^2-k^2b^2 &= (k^2-1)x^2-2(k^2b-a)x + (k^2-1)y^2 \\ && a^2-k^2b^2&= (k^2-1)\left(x-\frac{k^2b-a}{k^2-1}\right)^2-(k^2-1)\left(\frac{k^2b-a}{k^2-1}\right)^2+(k^2-1)y^2 \\ && \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2&= \left(x-\frac{k^2b-a}{k^2-1}\right)^2+y^2 \\ \Rightarrow && -7 &= \frac{k^2b-a}{k^2-1} \tag{*} \\ && 100 &= \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2 \\ &&&= \frac{a^2-k^2b^2}{k^2-1}+7^2 \\ \Rightarrow && 51 &= \frac{a^2-k^2b^2}{k^2-1} \tag{**} \\ (*) \Rightarrow && k^2(b+7)&= a+7 \\ (**) \Rightarrow && k^2(51+b^2)&= a^2+51 \\ \Rightarrow && \frac{a^2+51}{b^2+51} &= \frac{a+7}{b+7} \\ \\ \Rightarrow && a^2b+51b+7a^2 &= ab^2+51a+7b^2 \\ && 0 &= ab(b-a)-51(b-a)+7(b-a)(b+a) \\ &&&= (b-a)(ab+7(b+a)-51) \\ &&&= (b-a)((a+7)(b+7)-100) \\ \Rightarrow && 100 &= (a+7)(b+7) \end{align*} Since \(a \neq b\)

View