Question

Normal at $(5,3)$ of rectangular hyperbola $xy-y-2x-2=0$ intersects it again at a point

• A. $(\frac{3}{4},-14)$
• B. $(-1,0)$
• C. $(-1,1)$
• D. $(0,-2)$

Answer 1

The correct option is A $(\frac{3}{4},-14)$

$xy-y-2x-2=0\Rightarrow (x-1)(y-2)=4$
$\Rightarrow XY=4$
Normal at $(ct,\frac{c}{t})$ intersect it again at $(c{t}^{\prime },\frac{c}{t^{\prime }})$ then we know that $t^{\prime }=\frac{-1}{t^3}$
$x-1=X=ct$
$\Rightarrow 2t=4$
$\therefore t=2$
Hence $t^{\prime }=-\frac{1}{8}$
The intersection point will be $(X^{\prime },Y^{\prime })$
$=(\frac{-1}{4},-16)$
$\therefore (x^{\prime },y^{\prime })=(\frac{3}{4},-14)$