# Question 3057c

##### 1 Answer
Jul 5, 2016

$x + y \pm 1 = 0$

#### Explanation:

Consider the hyperbola

$f \left(x , y\right) = 3 {x}^{2} - 4 {y}^{2} - 12 = 0$

and the line

$g \left(x , y\right) = a x + b y + c = 0$

at tangency points $\left\{{x}_{0} , {y}_{0}\right\}$ the point normals are aligned

$\nabla f \left({x}_{0} , {y}_{0}\right) + \lambda \nabla g \left({x}_{0} , {y}_{0}\right) = \vec{0}$

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

${x}_{0} = - \frac{4 c}{a} , {y}_{0} = \frac{3 c}{a} , \lambda = \frac{24 c}{a} ^ 2$

Parameter $c$ is obtained substituting the found values in

$f \left({x}_{0} , {y}_{0}\right) + \lambda g \left({x}_{0} , {y}_{0}\right) = 0$ giving two solutions

$c = \pm a$

so the tangent lines are

$a x + a y \pm a = 0$ or

$x + y \pm 1 = 0$