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
JI
Related Maths A Level answers
All answers ▸How to perform integration by substitution. (e.g. Find the integral of (2x)/((4+(3(x^2)))^2)) (10 marks)
Differentiate with respect to x: F(x)=(x^2+1)^2
A particle A rests on a smooth inclined plane, it is connected to a particle B by a light inextensible string that is passed over a fixed smooth pulley at the top of the plane. B hangs freely. Find the acceleration of the system and tension in the string.
Solve the differential equation: dy/dx = 6x^2 + 4x + 9