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) = 4or y=2+4x−1dydx=−4(x−1)2Any point on the hyperbola is 1+2t, 2+2t=(5, 3)⇒t=2So slope of the normal at (5, 3) is−dxdy(5, 3)=4Equation of the normal at (5, 3) is y – 3 = 4(x – 5) which will pass through1+2t,2+2t if 2+2t=4(1+2t)−17⇒8t2−15t−2=0⇒t=2,−18.For t=−18, we get the required point as1−28, 2−21/8=34,−14