Question Number 103751 by hardylanes last updated on 17/Jul/20

  given that the parametric equation of   a curvd are x=(1/(t−1))  y=(1/(t+1)) obtaun a   cartesian equation of the curve. Hemce find  an equqtion of tbe nirmal to the curve at the   point t=2

Answered by bobhans last updated on 17/Jul/20

x = (1/(t−1)) →t−1 = (1/x); t+1 = (1/x)+2  y = (1/(t+1)) = (x/(1+2x)) . ⇒y′ = (1/((2x+1)^2 ))  slope normal line ⇒m = −(1/(y′)) = −(2x+1)^2   at point (x,y)=(1,(1/3)) ⇒m = −9  eq of the normal line ⇒9x+y = 9.1+1.(1/3)  9x+y = ((28)/3) or 27x + 3y = 28 ⊕

Commented bybobhans last updated on 17/Jul/20

what do you meant ???