Question

If $x+4y=14$ is a normal to the curve $y^2=\alpha x^3-\beta$ at point $(2,3),$ then value of $\alpha +\beta$ is

• A. $9$
• B. $-5$
• C. $7$
• D. $-7$

Answer 1

The correct option is A $9$

$y^2=\alpha x^3-\beta \Rightarrow \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$
$\Rightarrow$ slope of the normal at $(2,3)$ is
${(-\frac{dx}{dy})}_{(2,3)}=-\frac{2\times 3}{3\alpha \left(2^2\right)}$
$=-\frac{1}{2\alpha }=-\frac{1}{4}$
$\Rightarrow \alpha =2$
Also, $(2,3)$ lies on the curve.
$\Rightarrow 9=8\alpha -\beta$
$\Rightarrow \beta =16-9=7$
$\therefore \alpha +\beta =2+7=9$