If 3x + y = 8 is a tangent to the curve y^2=α+βx^3 at (2, 2), then va…

$3x + y = 8$

$y=-3x+8$

slope = $y'=-3$

$y^2=α+βx^3$

point (2, 2)

differentiating eq. w.r.t. x

$2y\frac{dy}{dx}=3βx^3$

at (x, y) = (2, 2) $\frac{dy}{dx}=-3$

$2(2)(-3)=3β(2)^2$

$=-3=3β⇒β=-1$  ...(1)

putting (2, 2) in original eq.

$2^2=α+β(2)^3$

from (1) $β=-1⇒4=α-8$

so $α=12$

$α-β=12+1=13$