Three normals are drawn from a point $P(1,1)$ to a parabola and let $A,B,C$ be feet of perpendiculars with $A=(0,0)$ and $B=(3,-1),$ find the slope of directrix. If equation of parabola is$ (x-a)^2+(y+b)^2=\frac{(x+cy-8)^2}{10}$, find $a+2b+c$
My Attempt:
Sum of ordinates of feet of normals is zero. Therefore, let $C=(h,1)$.
Also, second part of the question implies focus is $(a,-b)$ and directrix is $x+cy-8=0$.
A solution exists on toppr website but I don't understand how they calculated $a$.
I know the semi-latus rectum is the harmonic mean of the segments of focal chord but don't know how to use that here.
Also, is there any other way solve this question?
Edit: In the answer below, are we rejecting $c=3$? Why?
Best Answer
As written in Vedaansh Agarwal's answer, we have $$c^2=9\tag1$$ $$a^2+b^2=\frac{32}{5}\tag2$$ $$(3-a)^2+(-1+b)^2=\frac{(3-c-8)^2}{10}\tag3$$
Here, Vedaansh Agarwal wrote
However, this is not true.
For $c=3$, the parabola has a point $C(h,1)$. See here.
So, this means that we need another condition to eliminate $c=3$.
Differentiating the both sides of $$(x-a)^2+(y+b)^2=\frac{(x+cy-8)^2}{10}$$ we get $$2(x-a)+2(y+b)y'=\frac{(x+cy-8)(1+cy')}{5}$$
Substituting $(x,y,y')=(0,0,-1)$, we get $$-2a-2b=\frac{-8(1-c)}{5}\tag4$$
Solving $(1)(2)(3)(4)$ gives $$(a,b,c)=\bigg(\frac{12}{5},\frac 45,-3\bigg)$$
Therefore, the slope of directrix is $\color{red}{\frac{1}{3}}$ with $a+2b+c=\color{red}1$.