To find the equation of the normal to the given curve at the point (-1,1), you can use the following steps:
Rewrite the given equation in the form "y = mx + b," where m is the slope of the curve and b is the y-intercept.
Find the slope of the curve at the point (-1,1). This can be done by taking the derivative of the equation and evaluating it at x = -1.
The slope of the normal line at the point (-1,1) is the negative reciprocal of the slope of the curve at that point. Calculate this value.
Use the point-slope formula to write the equation of the normal line in the form "y - y1 = m(x - x1)," where (x1, y1) is the point (-1,1) and m is the slope of the normal line.
Substitute the values for x1, y1, and m into the point-slope formula to obtain the final equation of the normal line.
For example, if the given equation is 2x^2 + 2y^2 - 4x + 4y = 12, you can follow these steps:
Rewriting the equation in slope-intercept form, we get: y = -x + 2
Taking the derivative of the equation, we get: y' = -1
The slope of the normal line is the negative reciprocal of the slope of the curve, which is 1/-1 = -1
Using the point-slope formula, we get: y - 1 = -1(x + 1)
Substituting the values into the point-slope formula, we get: y - 1 = -1x - 1
Thus, the equation of the normal line at the point (-1,1) is y - 1 = -1x - 1.