The locus of the point of intersection of the lines $\left(\sqrt[]{3}\right)kx+ky-4\sqrt[]{3}=0and\sqrt[]{3}x-y-4\left(\sqrt[]{3}\right)k$ = 0 is a conic, whose eccentricity is…………………

$?$ ae = 2b

$\frac{4b^2}{a^2}=e^2$  

Or 4 (1 – e²) = e²

4 = 5e² -> $e=\frac{2}{\sqrt[]{5}}$

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| <  $\sqrt[]{9+16}$ < r + 2

|r – 2| < 5                     and r + 2 > 5

–5 < r – 2 < 5                    r > 3                      … (2)

–3 < r < 7                                                        … (1)

From (1) and (2)

3 < r < 7

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$ &n

Slope of axis = $\frac{1}{2}$

$y-3=\frac{1}{2}\left(x-2\right)$              

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

            α + 1.6 = 4 ⇒ α = 2.4

            β + 2.8 = 6 ⇒

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$       

= 2 cos² θ + 2 sin² θ + 6 sin θ + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin q = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.