Question #3057c

1 Answer
Jul 5, 2016

#x+y pm 1=0#

Explanation:

Consider the hyperbola

#f(x,y) = 3x^2-4y^2-12=0#

and the line

#g(x,y) = a x + by +c = 0#

at tangency points #{x_0,y_0}# the point normals are aligned

#grad f(x_0,y_0)+lambda grad g(x_0,y_0) = vec 0#

or

#{ (a lambda + 6 x=0), (b lambda - 8 y=0),( c + a x + b y=0) :}#

Solving for #x_0,y_0,lambda#

#x_0 = -(4 a c)/(4 a^2 - 3 b^2), y_0 = -( 3 b c)/( 3 b^2-4 a^2), lambda = (24 c)/(4 a^2 - 3 b^2)#

but we are interested in tangent lines which make equal intercepts on the axes. So this implies that #a = b# then the tangency points are

#x_0 = -(4 c)/a, y_0 = (3 c)/a, lambda = (24 c)/a^2#

Parameter #c# is obtained substituting the found values in

#f(x_0,y_0)+lambda g(x_0,y_0)=0# giving two solutions

#c = pm a#

so the tangent lines are

#ax + aypm a=0# or

#x+y pm 1=0#

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