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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