Given a point P (x, y) on the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0), find the value range of 3x + 4Y X * 2 is the square of X /Denotes divided by

You can try triangle substitution
X=a*sin(c) Y=b*cos(c)
3X+4Y=3a*sin(c)+4b*cos(c)
=SQR (9a ^ 2 + 16b ^ 2) * sin (x + D) auxiliary angle formula
So 3x + 4Y belongs to (- SQR (9a ^ 2 + 16b ^ 2), SQR (9a ^ 2 + 16b ^ 2))
SQR is open root

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