Question

Find the equation of tangents to the hyperbola $3x^2-4y^2=12$, which make equal intercepts on the axes.

Answer 1

Equation of the hyperbola is,

$3x^2-4y^2=12$ ........ (1)

Tangent of the hyperbola makes equal intercepts on the axes. So the tangent is of the form $x+y=a$

To find the point of contact of the tangent, substitute $y=a-x$ in equation (1).

$3x^2-4(a-x{)}^2=12$

$3x^2-4a^2-4x^2+8ax=12$

$x^2-8ax+4a^2+12=0$

Because the tangent will touch on a point, there will be single solution for the above Quadratic equation

$b^2-4ac=0$

$64a^2-4(4a^2+12)=0$

$48a^2-48=0$

$a=\pm 1$

So the tangents are $x+y=1$ and $x+y=-1$