Explanation

(a)
Given, $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ …(i)
$\Rightarrow \ \ \frac{2x}{3} - y\frac{dy}{dx} = 0$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{2x}{3y}$
Now, slope of the line $y = x$ is 1
Since, tangent is parallel to given line, then
$\frac{2x}{3y} = 1\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x = \frac{3y}{2}$ ...(ii)
From Eqs. (i) and (ii), we get $y = \pm 2$
$\therefore$ Points of contact are (3, 2) and $( - 3,\ - 2)$
$\therefore$ equations of tangents are
$y - 2 = (x - 3)\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ \ x - y - 1 = 0$
and $y + 2 = x + 3\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ \ x - y + 1 = 0$