Explanation

(a)
Given equation is $x^{2} + 9y^{2} - 4x + 3 = 0$ …(i)
or $x^{2} - 4x + 9y^{2} + 3 = 0$
Since $x$ is real.
$\therefore( - 4)^{2} - 4(9y^{2} + 3) \geq 0$
$\Rightarrow 16 - 4(9y^{2} + 3) \geq 0$
$\Rightarrow 4 - 9y^{2} - 3 \geq 0$
$\Rightarrow \ 9y^{2} - 1 \leq 0$
$\Rightarrow (3y - 1)(3y + 1) \leq 0$
$\Rightarrow \ \frac{- 1}{3} \leq y \leq \frac{1}{3}$
Eq. (i) can also be written as
$9y^{2} + 0y + x^{2} - 4x + 3 = 0$
Since $y$ is real.
$\therefore 0^{2} - 4.9(x^{2} - 4x + 3)) \geq 0$
$\Rightarrow x^{2} - 4x + 3 \leq 0$
$\Rightarrow (x - 1)(x - 3)) \leq 0$
$\Rightarrow 1 \leq x \leq 3$