Parabola

Given, $y=m_1 x+c_1$ is common tangent of circle and parabola.

For circle, $\left|\frac{c_1}{\sqrt{1+m_1^2}}\right|=\sqrt{2} \Rightarrow c_1^2=2\left(1+m_1^2\right)$ .... (i)

For parabola, $c_1=\frac{1}{4 m_1} \Rightarrow c_1^2=\frac{1}{16 m_1^2}$ .... (ii)

Also, $y=m_2 x+c_2$ is common tangent of circle and parabola.

For circle, $\left|\frac{c_2}{\sqrt{1+m_2^2}}\right|=\sqrt{2} \Rightarrow c_2^2=2\left(1+m_2^2\right)$

For parabola, $c_2=\frac{1}{4 m_2} \Rightarrow c_2^2=\frac{1}{16 m_2^2}$

Now, $2\left(1+m_1^2\right)=\frac{1}{16 m_1^2} \quad$ [From Eqs. (i) and (ii)]

$\begin{aligned} & \Rightarrow \quad 32 m_1^4+32 m_1^2-1=0 \\ & \Rightarrow \quad m_1^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32} \end{aligned}$

$\Rightarrow \quad m_1^2=\frac{-32+24 \sqrt{2}}{64}$

[Ignoring negative value because $m^2 \geq 0$]

$\Rightarrow m_1= \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}$

Similarly, $m_2= \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}$

$\begin{aligned} \therefore 8\left|m_1 m_2\right| =8\left|\left( \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}\right)\left( \pm \sqrt{\frac{-32+24 \sqrt{2}}{64}}\right)\right| \\ =\frac{8}{64}(-32+24 \sqrt{2}) \\ =-4+3 \sqrt{2} \end{aligned}$

Solving the given equations,

${(mx + c)^2} = 4ax - 4{a^3}$

$ \Rightarrow {m^2}{x^2} + 2mc\,.\,x + {c^2} = 4ax - 4{a^3}$

$ \Rightarrow {m^2}{x^2} + (2mc - 4a)x + {c^2} + 4{a^3} = 0$

Since the straight line touches the parabola at a point, so, the discriminant = 0

$ \Rightarrow {(2mc - 4a)^2} - 4{m^2}({c^2} + 4{a^3}) = 0$

$ \Rightarrow 4{m^2}{c^2} - 16amc + 16{a^2} - 4{m^2}{c^2} - 16{a^3}{m^2} = 0$

$ \Rightarrow - mc + a - {a^2}{m^2} = 0$

$ \Rightarrow mc = a - {a^2}{m^2} \Rightarrow c = {a \over m} - {a^2}m$

For the parabola ${y^2} = 4ax$, the equation of normal at $P(a{m^2}, - 2am)$ is $y = mx - 2am - a{m^3}$.

Here, $m = \tan 60^\circ = \sqrt 3 $

$\therefore$ $P \equiv (a{(\sqrt 3 )^2}, - 2a(\sqrt 3 )) \equiv (6, - 4\sqrt 3 )$ ($\because$ $a = 2$)

Thus, P satisfies $2x - y - 12 - 4\sqrt 3 = 0$

Suppose the parabola, y = x² + px + q cuts X-axis at A($\alpha$, 0) and B($\beta$, 0).

Then, $\alpha$, $\beta$ are roots of the equation x² + px + q = 0

$\therefore$ $\alpha$ + $\beta$ = $-$p and $\alpha \beta$ = q

The parabola y - x² + px + q cuts Y-axis at (0, q).

Let the equation of the circle passing through A, B and C be

x² + y² + 2gx + 2fy + c = 0 .... (i)

$\therefore$ $\alpha$² + 2g$\alpha$ + c = 0 ..... (ii)

$\beta$² + 2g$\beta$ + c = 0 ...... (iii)

and q² + 2fq + c = 0 ..... (iv)

Subtracting Eq. (iii) from Eq. (ii), we get

$\alpha + \beta + 2g = 0$

$\Rightarrow$ g = p/2

Adding Eqs. (ii) and (iii), we get

$\alpha^2+\beta^2+2g(\alpha+\beta)+2c=0$

$(\alpha+\beta)^2-2\alpha \beta+2g(\alpha+\beta)+2c=0$

p² $-$ 2q $-$ p² + 2c = 0 [$\because$ $\alpha+\beta=p$ and g = $\frac{p}{2}$]

Putting c = q in Eq. (iv), we get f = $-$$\left(\frac{q+1}{2}\right)$

Substituting the values of g, f and c in Eq. (i), we obtain the equation of family of circles passing through A, B and C as

x² + y² + px $-$ (q + 1) y + q = 0

Clearly, it passes through (0, 1).

Let P(x, y) be the original position of the point w.r.t. the original axes. Let us move the origin at new position to (h, k).

Hence, the position of the same point P in the new system is

x' = x $-$ h $\Rightarrow$ y' = y $-$ k

Here, (h, k) = (1, 2)

$\therefore$ x' = (x $-$ 1), y' = (y $-$ 2)

According to the question,

y² $-$ 8x $-$ 4y + 12 = (y $-$ 2)² $-$ 4a(z $-$ 1)

$\Rightarrow$ y² $-$ 8x $-$ 4y + 12 = y² $-$ 4y + 4 $-$ 4ax + 4a

On comparing respective coefficients, we get

4a = 8 $\Rightarrow$ a = 2

Given equation of parabola is

x = 4y $-$ y²

$\Rightarrow$ y² $-$ 4y + 4 = $-$ x + 4

$\Rightarrow$ (y $-$ 2)² = $-$ (x $-$ 4)

a = ${{1} \over 4}$

Focus = $\left( {{{15} \over 4},2} \right)$

Parabola cuts Y-axis of (0, 0) and (0, 4)

Now, equation of tangent at (0, 0) is

x = 4y ...... (i)

and equation of tangent at (0, 4) is

x + 4y = 16 .... (ii)

Solving Eqs. (i) and (ii), we get

x = 8, y = 2

$\therefore$ Intersection point of tangent is P(8, 2).

Distance from point P to focus S

Distance = $\sqrt {{{\left( {8 - {{15} \over 4}} \right)}^2} + {{(2 - 2)}^2}} $

$ = 8 - {{15} \over 4} = {{17} \over 4}$