If the line $6 x-y-4=0$ touches the curve $y^{2}=a x^{3}+b$ at the point $(1,2)$ then $a+b=$
If the line $6 x-y-4=0$ touches the curve $y^{2}=a x^{3}+b$ at the point $(1,2)$ then
$a+b=$
8
$-4$
4
12
Solution
Slope of line $6 x-y-4=0$ is 6 and this line is tangent to the curve $y^{2}=a x^{3}+b$ at point $(1,2)$
$\therefore 2 y \frac{d y}{d x}=3 a x^{2} \Rightarrow\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{3 a x^{2}}{2 y}=\frac{3 a}{4}$ and $\frac{3 a}{4}=6 \Rightarrow a=8$
Now point $(1,2)$ lies on given curve.
$\therefore(2)^{2}=(8)(1)^{3}+b \Rightarrow b=-4 \Rightarrow a+b=8-4=4$