$d = \sqrt{(x - 4)^2 + (y - 2)^2}$

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*