Normal at point (5, 3) to the rectangular hyperbola $xy – y – 2x – 2 = 0$

meets the curve at the point whose coordinates are

Equation of the hyperbola can be written as $(x – 1) (y – 2) = 4$

or $y=2+\frac{4}{x-1}$

$\frac{dy}{dx}=-\frac{4}{(x-1{)}^2}$

Any point on the hyperbola is 

$\left(1+2t, 2+\frac{2}{t}\right)=(5, 3)\Rightarrow t=2$

So slope of the normal at (5, 3) is

$-{\left.\frac{dx}{dy}\right|}_{(5, 3)}=4$

Equation of the normal at (5, 3) is y – 3 = 4(x – 5) which will pass through

$\left(1+2t,2+\frac{2}{t}\right)\text{ if }2+\frac{2}{t}=4(1+2t)-17$

$\Rightarrow 8t^2-15t-2=0\Rightarrow t=2,-\frac{1}{8}$.

For $t=-\frac{1}{8},$ we get the required point as

$\left(1-\frac{2}{8}, 2-\frac{2}{1/8}\right)=\left(\frac{3}{4},-14\right)$