Final answer:
To find the equation of the normal to the curve y = sqrt(x) from an external point (4,0), we calculate the derivative to find the slope of the tangent at any point on the curve, then find the negative reciprocal of that slope for the normal. The slope of the normal is -2sqrt(x), and we use the point-slope form to find the specific equation, considering the given point and the slope.
Step-by-step explanation:
To find the equation of the normal to the curve y = √ x at a point P from an external point (4,0), we first need to determine the coordinate of P where the normal will pass. To find this point, we must calculate the derivative of the curve to find the slope of the tangent at any point on the curve, and then find the slope of the normal, which will be perpendicular to this tangent.
The derivative of y = √ x is dy/dx = 1/(2√ x). The slope of the normal is the negative reciprocal of dy/dx, which is -2√ x. Since the normal passes through the point (4,0), we can set up the equation of the normal using the point-slope form: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point through which the line passes.
Plugging in the given point and the slope, we get the equation: y - 0 = -2√ x (x - 4). We then need to find the x coordinate of P where this line is normal to the curve, potentially by setting the slope equal to the negative reciprocal of the derivative at P.
After determining the x-coordinate, we substitute it back into our normal equation to find the specific equation of the normal line we're looking for.