Since the vertices are (0, ±7), equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
a = 7 and e = \(\frac { 4 }{ 3 } \)
b² = a²(e² - 1) = 49\(\left( \frac { 16 }{ 9 } -1 \right) =49\left( \frac { 16-9 }{ 9 } \right) \)
= \(49\left( \frac { 7 }{ 9 } \right) =\frac { 343 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 49 } -\frac { { x }^{ 2 } }{ \frac { 343 }{ 9 } } =1\) 
⇒ \(\frac { { y }^{ 2 } }{ 49 } -\frac { 9{ x }^{ 2 } }{ 343 } =1\)

Let the equation of the ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given, length of LR = \(\frac { 1 }{ 2 } \) (Length of major axis)
⇒ \(\frac { { 2b }^{ 2 } }{ a } =\frac { 1 }{ 2 } \) (2a) ⇒ \(\frac { 2{ b }^{ 2 } }{ a } \) = a ⇒ 2b² = a²
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { { b }^{ 2 } }{ { 2a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { 1 }{ \sqrt { 2 } } \)

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Let AOB be the vertical section of the reflector and m is the mid-point of AB. Let the equation of the parabola be y² = 4ax A(6, 12) lies on (1)

∴ 12² = 4a(6) ⇒ a = 6
∴ Focus is (a, 0) = (b, 0)
Hence focus coincides with m, the mid-point of AB.

Given circle is x² + y² - 8x - 9y + 12 = 0
Length of the tangent = \(\sqrt { { 2 }^{ 2 }+({ -3) }^{ 2 }-8(2)-9(-3)+12 } \)
= \(\sqrt { 4+9-16+27+12 } \)
= \(\sqrt { 36 } \) = 6 unit