How do you find an equation for the tangent line to #x^4=y^2+x^2# at #(2, sqrt12)#?

2 Answers
Jan 18, 2017

#y=7/sqrt3x-8/sqrt3# or #sqrt3y=7x-8#

Explanation:

#x^4=y^2+x^2#
#y^2=x^4-x^2#
#2y(dy/dx)=4x^3-2x#
#dy/dx=(4x^3-2x)/(2y)#

at #(2,sqrt12)#
#dy/dx=(4(2^3)-2(2))/(2sqrt12)#
#dy/dx=(32-4)/(4sqrt3)#
#dy/dx=28/(4sqrt3)#
#dy/dx=7/sqrt3#

The equation of tangent, #y-y_1=m(x-x_1)# where #m=7/sqrt3, y_1=sqrt12 and x_1=2#

#y-sqrt12=7/sqrt3(x-2)#
#y-sqrt12=7/sqrt3x-14/sqrt3#
#y=7/sqrt3x-14/sqrt3+sqrt12#
#y=7/sqrt3x-14/sqrt3+sqrt36/sqrt3#
#y=7/sqrt3x-14/sqrt3+6/sqrt3#
#y=7/sqrt3x-8/sqrt3# or
#sqrt3y=7x-8#

Jan 18, 2017

#7x-sqrt3y-8=0#. See tangent-inclusive graph. The graph is not to scale. There is contraction in the y-direction.

Explanation:

Differentiating,

#4x^3=2x+2yy'#, giving #y'=7/sqrt3#, at #P(2, sqrt12)#.

The equation to the tangent at P is

y-sqrt13=7/sqrt3(x-2), giving

#7x-sqrt3y-8=0#.

graph{(x^2-sqrt(x^2+y^2))(7x-sqrt3 y-8.2)=0 [-7, 7, -35, 35]}