
$d = \sqrt{(x - 4)^2 + (y - 2)^2}$
from
$x^2 + 3y^2 = 12$
$x = \sqrt{12 - 3y^2}$
$d = \sqrt{\left( \sqrt{12 - 3y^2} - 4 \right)^2 + (y - 2)^2}$
$\dfrac{dd}{dy} = \dfrac{2\left( \sqrt{12 - 3y^2} - 4 \right) \dfrac{-6y}{2\sqrt{12 - 3y^2}} + 2(y - 2)}{2\sqrt{\left( \sqrt{12 - 3y^2} - 4 \right)^2 + (y - 2)^2}} = 0$
$-6y + \dfrac{24y}{\sqrt{12 - 3y^2}} + 2y - 4 = 0$
$\dfrac{24y}{\sqrt{12 - 3y^2}} = 4y + 4$
$\dfrac{6y}{\sqrt{12 - 3y^2}} = y + 1$
$\dfrac{36y^2}{12 - 3y^2} = (y + 1)^2$
$\dfrac{36y^2}{3(4 - y^2)} = (y + 1)^2$
$12y^2 = (y + 1)^2 (4 - y^2)$
$12y^2 = (y^2 + 2y + 1) (4 - y^2)$
$12y^2 = 4y^2 + 8y + 4 - y^4 - 2y^3 - y^2$
$y^4 + 2y^3 + 9y^2 - 8y - 4 = 0$
By trial and error
$y = 1$
$x = \sqrt{12 - 3(1^2)} = 3$
The nearest point is (3, 1)
Nearest distance:
$d = \sqrt{(3 - 4)^2 + (1 - 2)^2}$
$d = \sqrt{2}$ answer