How can I find the normal to a curve at a given point?
Take the curve y = 2x2 - 4 and point P (2,4). a) First, find the straight line tangent to the curve at point P. This will have the gradient of the curve at point P. To find it, differentiate: dy/dx = 4x. At P, this will equal 4(2) = 8. As the tangent has a gradient of 8 and passes through P, its equation can be found using dy/dx = (y2 - y1)/(x2 - x1), where (x2 ,y2) are the coordinates of P and (x1 ,y1) are the coordinates of the y-intercept. We know x1 = 0 since the y-intercept is on the y-axis. We need to find y1, so: 8 = (4 - y1)/(2 - 0); 8 * 2 = 4 - y1; 4 - 16 = y1; y1 = -12. The equation of the tangent is y = 8x - 12. b) To find the normal, we need to find its gradient and y-intercept. The gradients of perpendicular lines are inverse. Therefore, the gradient of the normal is 1/8. We use the same method as previously to find the y-intercept: 1/8 = (4 - y1)/(2 - 0); 1/8 * 2 = (4 - y1); 4 - 1/4 = y1; y1 = 3.75. So, the equation of the normal to the curve at point P is y = 1/8x + 3.75.
