For a graph C with equation y=3/(5-3x)^2, find the the equation of the line normal to the graph at point P, where x=2. Give your answer in the form ax+by+c=0

Find out the value of the y coordinate where x=2y=3/(5-3(2))^2=3Differentiate function in order to find the gradient of the graph at point Py=3/(5-3x)^-2Use chain rule where u=5-3xdu/dx=-3dy/dx=3u^-2Hence dy/dx=-6u^-3*-3dy/dx=18(5-3x)^-3dy/dx at x=2 is 18(5-3*2)^-3=-18Take negative reciprocal of 18 to find gradient of line normal to graph-18-->1/18y-y1=m(x-x1)Substitute in valuesy-3=1/18(x-2)18y-54=x-218y-x-52=0-x+18y-52=0