The sides of a triangle have the combined equation `x^2-3y^2-2x y+8y-4=0` . The third side, which is variable, always passes through the point `(-5,-1)` . Find the range of values of the slope of the third line such that the origin is an interior point of the triangle.

Answer

4

Explanation

`x^(2) - 3y^(2) - 2xy + 8y - 4 - 0 equiv (x - 3y + 2)(x + y - 2)`
note that `(-5, -1)` lies on `x - 3y + 2 = 0`.
In limiting case line passing through
`(-5, 1)` can be `||` to `x + y - 2 = 0`
i.e. `m gt - 1`
and maximum slope can occur if it passes through `(0, 0)` i.e., `m lt (1)/(5) rArr m in (-1, (1)/(5)) rArr a = -1` and `b = (1)/(5)`
Hence `(a + (1)/(b)) = -1 + 5 = 4`