2003 Paper 3 Q7

Year: 2003
Paper: 3
Question Number: 7

Course: UFM Pure
Section: Conic sections

Difficulty: 1700.0 Banger: 1484.0

conics perpendicular locus geometric proof coordinate geometry line equations circle right angle

Problem

In the $x$--$y$ plane, the point $A$ has coordinates $(a\,,0)$ and the point $B$ has coordinates $(0\,,b)\,$, where $a$ and $b$ are positive. The point $P\,$, which is distinct from $A$ and $B$, has coordinates~$(s,t)\,$. $X$ and $Y$ are the feet of the perpendiculars from $P$ to the $x$--axis and $y$--axis respectively, and $N$ is the foot of the perpendicular from $P$ to the line $AB\,$. Show that the coordinates $(x\,,y)$ of $N$ are given by \[ x= \frac {ab^2 -a(bt-as)}{a^2+b^2} \;, \ \ \ y = \frac{a^2b +b(bt-as)}{a^2+b^2} \;. \] Show that, if $\ds \ \left( \frac{t-b} s\right)\left( \frac t {s-a}\right) = -1\;$, then $N$ lies on the line $XY\,$. Give a geometrical interpretation of this result.

No solution available for this problem.

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Problem source

In the $x$--$y$ plane, the point $A$ has coordinates    
$(a\,,0)$ and the point $B$ has coordinates $(0\,,b)\,$,    
where $a$ and $b$ are positive.    
The point $P\,$, which is distinct from $A$ and $B$, has coordinates~$(s,t)\,$.
$X$ and $Y$ are the feet of the perpendiculars from $P$ to the $x$--axis and
$y$--axis respectively, and     
$N$ is the foot of the perpendicular from $P$ to the line $AB\,$.   
Show that the coordinates $(x\,,y)$ of $N$ are given by   
\[   
x= \frac {ab^2 -a(bt-as)}{a^2+b^2} \;, \ \ \    
y = \frac{a^2b +b(bt-as)}{a^2+b^2} \;.   
\]   
   
Show that, if    
$\ds   \
\left( \frac{t-b} s\right)\left( \frac t {s-a}\right) = -1\;$, then $N$ lies on    
the line $XY\,$.   
   
Give a geometrical interpretation of this result.

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