The sum of the y -intercepts of the tangents drawn from the point...

Let, the equation of the tangent from $\left(- 2, - 1\right)$ is $y = mx + \sqrt{3{m}^{2} - 2}$
$⇒ - 1 = - 2m + \sqrt{3{m}^{2} - 2}$
$⇒ {\left(2m - 1\right)}^{2} = 3{m}^{2} - 2$
$⇒ {m}^{2} - 4m + 3 = 0$
$⇒ m = 1, 3$
So, the equations of the tangents are
\left(y + 1\right) = \left(x + 2\right) \text{ and } \left(y + 1\right) = 3\left(x + 2\right)$<br />$⇒ y$-intercepts of the tangents are$1&5$<br />$⇒$the sum of$y$-intercepts$= 6