Question

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

Answer 1

$y^2=\alpha x^3-\beta$
Since $(2,3)$ lies on the curve,
$9=8\alpha -\beta \cdots \left(1\right)$

$y^2=\alpha x^3-\beta$
Differentiating w.r.t. $x,$
$2y{y}^{\prime }=3\alpha x^2$
At point $(2,3),$ $y^{\prime }=2\alpha$
$\therefore$ Slope of normal at $(2,3)$ is $-\frac{1}{2\alpha }$
Given $x+4y=14$ is a normal to the curve $y^2=\alpha x^3-\beta$ at $(2,3)$
Slope $=-\frac{1}{4}$
$-\frac{1}{2\alpha }=-\frac{1}{4}\Rightarrow \alpha =2$
Putting $\alpha =2$ in eqn.$\left(1\right),$ we get $\beta =7$
$\therefore \alpha +\beta =9$