How do you find the equations for the normal line to #x^2+y^2=20# through (2,4)?

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Answer 1 · 2017-04-23T03:49:42

#y=2x#

#y-0 = (4-0)/(2-0)(x-0)#

#y=2x#

:)))

Answer 2 · 2017-04-23T03:55:38

The equation is #y =2x#.

The derivative of this relation is given by

#2x + 2y(dy/dx) = 0#

#2y(dy/dx) = -2x#

#dy/dx= -x/y#

The slope of the tangent line is

#m_"tangent" = -2/4 = -1/2#

Therefore, the slope of the normal line is #2#.

#y - 4 = 2(x - 2)#

#y- 4= 2x - 4#

#y = 2x#, as obtained in the other answer

Hopefully this helps!