Practicing Success
If $3x + y = 8$ is a tangent to the curve $y^2=α+βx^3$ at (2, 2), then value of $α-β$ is |
12 -1 11 13 |
13 |
$3x + y = 8$ $y=-3x+8$ 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 $α=12$ $α-β=12+1=13$ |