(derivative) Find the equations of the normal lines to the graph of x^2 + 2y^2 = 36 that pass through point (1?

A line through (1, 0) with slope m is given by y = m(x - 1).
Find where this intersects with the ellipse, for non-zero m.
The product of m and the slope of the tangent to the ellipse at the point of intersection must be -1. Use this to find possible values of m.

Use the formula y = ax + b the place a is the slope... the slope of the given line is -a million/3 and for a perpendicular the slope could desire to be 3 Then it is y = 3x+b ... now use the factor to discover b ... it is 7 = 3*2 + b ==> b =a million Then the equation is y = 3x + a million ok!

2x+4ydy/dx=0 dy/dx=-x/2y (1,0) =-1/inf slope is 0