Find the equations of the normal lines to the graph of x^2 + 2y^2 = 36 that pass through point (1,0).
I found one, that is y = 0, but the book says y = 4x - 4 and y = -4x + 4 are answers too. I can't find the other two. How do you solve this question?
THX!
(derivative) Find the equations of the normal lines to the graph of x^2 + 2y^2 = 36 that pass through point (1?
2011-11-02 12:39 pm
回答 (3)
2011-11-02 1:19 pm
✔ 最佳答案
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.
2016-12-10 6:57 pm
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!
2011-11-02 12:52 pm
2x+4ydy/dx=0 dy/dx=-x/2y (1,0) =-1/inf slope is 0
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