Norml, Application of derivatives

Question

If $x+4y =14$ is normal to the curve $y^2=αx^3 - β$ at $(2,3)$, then the value of $α+β$ is? I equated the slope of the normal with the value of $-dx/dy$ and found $α=2$, how do I find $β$?

Answer 1

Since $(2,3)$ is a point of the curve, then $3^2=\alpha\times2^3-\beta$. In other words, $\beta=8\alpha-9$. So, your curve is $y^2=\alpha x^3-8\alpha+9$. Let $F(x,y)=y^2-\alpha x^3$. Then $\nabla F(2,3)=(-12\alpha,6)$. So, the slope of the normal is $-\frac1{2\alpha}$. But then $-\frac1{2\alpha}=-\frac14$. So, $\alpha=2$ and $\beta=7$.

Answer 2

The slope of the normal line is given by $$m_N=-\frac{1}{4}$$ The slope of the curve is given by $$2y'y=3\alpha x^2$$ Can you finish? and we get $$-\frac{1}{4}=-\frac{2y}{3\alpha x^2}$$