Question

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

• A. $9$
• B. $-5$
• C. $7$
• D. $-7$

Answer 1

The correct option is A $9$

Given equation of curve is
$y^2=\alpha x^3-\beta$
Since, it passes through (2,3)
$4=8\alpha -\beta$
$\frac{dy}{dx}=\frac{3\alpha x^2}{2y}$
Slope of tangent at (2,3) is $2\alpha$
Slope of normal at $(2,3)=-\frac{1}{2\alpha }$
Given equation of normal is $x+4y=14$
Slope of normal is $-\frac{1}{4}$
$\Rightarrow \alpha =2$
$\Rightarrow \beta =7$
Hence $\alpha +\beta =9$