Problem source

\begin{questionparts}
\item 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
\[
\displaystyle \l x+7 \r^2 + y^2 =100 \;.
\]
\item 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 $(a+7)(b+7)=100\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item 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*}

\item 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 $a \neq b$


\end{questionparts}