Find the least positive value of ordinate of $C$ Let $A(0,2),B$ and $C$ are points on curve $y^2=x+4,$ and such that $\angle CBA=\frac{\pi}{2}.$Then find the least positive value of ordinate of $C$. * * * $y^2=x+4$ is a equation of a parabola whose vertex is $(-4,0)$ and the point $A$ is on the $y-$axis.If we take $B$ as $(0,-2)$ and $C$ as $(-4,0)$ ,then $BC$ is perpendicular to $AC$.But i dont know how to find the least positive value of ordinate of $C$. Please help me.

Let $B(b^2-4,b),C(c^2-4,c)$ where $b\
ot=\pm 2$.

Then, the line (call it $L$) that passes through $B$ and is perpendicular to the line $AB$ is given by $$L : y-b=-\frac{b^2-4}{b-2}(x-(b^2-4))\iff y=-(b+2)x+K$$ where $K=-(b+2)(-b^2+4)+b$.

Here, note that the intersection point of $L$ with $y^2=x+4$ other than $B$ is $C$. Since we have $$y^2=\frac{y-K}{-(b+2)}+4\iff (b+2)y^2+y-K-4(b+2)=0,$$ by Vieta's formulas, we have $$b+c=-\frac{1}{b+2}\iff c=-b-\frac{1}{b+2}.$$ So, we want to find least positive value of $-b-\frac{1}{b+2}$ for $b\
ot=\pm 2$.

So, the answer will be $\color{red}{4}$ (for $b=-3$).