Question involving normal and directrix of parabola

Question

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?

Answer 1

Given that

$(x-a)^2 + (y+b)^2= \frac{(x+cy-8)^2}{10}$

Taking root both sides and converting it in the form of SP = e*SM (where e is 1 for parabola)

$\sqrt{(x-a)^2+ (y+b)^2}= \frac{(x+cy-8)}{\sqrt10}$

We see that focus is (a,-b) and directrix is x+cy-8=0, as noted by you.

Also, the RHS is the perpendicualr distance of a point from the directrix. Comparing it with

$\frac{ah + bk+ c}{\sqrt{a^2+b^2}}$

[perpendicular distance of (h,k) from the line ax+by+c=0]

$\sqrt{c^2+1} = \sqrt10$

This gives $c^2 = 9$

Also, slope of directrix x+cy-8=0 is $\frac{-1}{c}$ by comparing with y= mx +c

Also, putting (0,0) and (3,-1) in the equation you get

$a^2 + b^2 = 6.4$

$a^2+b^2-6a-2b+10=\frac{c^2+10c+10}{10}$

Write the second equation in terms of $a^2+b^2$ and set equal to first equation. Simplifying, and putting c as -3 we get

3a+b=2 b=2-3a

Substitute in eq 1. We get a= $\frac{12}{5}, b= \frac{4}{5}$

Hence, a+2b+c = 1

Why not +3?

To answer that, I did the same but took c as +3. You will get a bit complicated values for a and b. But when you put these values in the parabola's equation, there is no point with the y coordinate 1. Hence, this parabola will not have a point C(h,1).

Answer 2

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

Why not +3? To answer that, I did the same but took c as +3. You will get a bit complicated values for a and b. But when you put these values in the parabola's equation, there is no point with the y coordinate 1. Hence, this parabola will not have a point C(h,1).

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 $a+2b+c=\color{red}1$.

Answer 3

Note that the normals at $A,B$ are perpendicular and hence the tangents also are perpendicular. Let them meet at $Q$. So $Q,A,B,P$ is a rectangle where diagonals bisect each other at $M \left(\dfrac{3}{2}, -\dfrac{1}{2} \right)$. So slope of $PM$ = slope of $QM$ = $-3$ and its well known that slope of $QM =$ slope of axis. From this we see that slope of directrix is $-\dfrac{1}{3}$

The tangents meet at $Q(2,-2)$ which lies on the directrix and hence equation of directrix is $x-3y-8=0$. Its now clear from the given equation that the focus is $(a,-b)$ and that $c=-3$.

To obtain the focus we first look at foot of perpendicular from $A(0,0)$ onto the directrix which is $\left(\frac{4}{5}, -\frac{12}{5} \right)$

This also happens to be the reflection of focus across the tangent $QA \equiv x+y=0$. Thus the focus is obtained as $S\left( \frac{12}{5},-\frac{4}{5}\right)$ so that $a=\frac{12}{5}, b = \frac{4}{5}$