An approach with a bit of geometry:
$3x^2 + 4xy + 6y^2 = 140$ is an ellipse rotated about the origin.
$3x^2 + 4xy + 6y^2 = 140$;
$2x^2 + (x+2y)^2 + 2y^2 = 140$;
$(x^2 + y^2) = 140/2 - [(x + 2y)^2]/2$.
Let $r$ be the distance from the origin,
$r^2 = (x^2 + y^2) = 70 - $
$[(x + 2y)^2 ]/2$.
Since the bracket on the r.h.s., a square, is $\\ \geq 0$, we get for the maximum:
$r^2 = 70$, at $ 2y + x = 0$.
The major axis of the ellipse, call it $a$, lies along the line $2y + x = 0$, and has squared length:
$a^2 = 70$.
The major axis lies along $2y +x = 0$, or $y = - (1/2)x$ , a straight line with slope $- 1/2$.
The line $y = 2x$ is perpendicular to $y = - (1/2) x$, the orientation of the major axis, I.e. a line along the minor axis.
Intersection of $y = 2x$ with the ellipse:
$3x^2 + 4x(2x) + 6(2x)^2 = 140$;
$35x^2 = 140$;
$\\x^2 = 4$, and using $y= 2x$, we find
$y^2 = 4x^2 = 16$.
$r^2 = min ( x^2 +y^2) = 20$;
For the squared minor axis, call it $b$, we have
$b^2 = 20$.