x²+3y² = 12
\(\div 12\) we get, \(\frac { { x }^{ 2 } }{ 12 } +\frac { { y }^{ 2 } }{ 4 } =1\)
∴ a² = 12, b² = 4
The line x-y+ 4 = 0 can be rewritten as y = x+4.
∴ m = 1, c = 4
The condition for y = mx + 4 to be a tangent to the ellipse is c² = a²m² + b²
∴ (4)² = 12(1)² + 4
⇒ 16 = 12+4
⇒ 16 = 16
Since the condition is satisfied, the line x - y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12.
Also, the point of contact is \(\left( -\frac { { a }^{ 2 }m }{ c } ,\frac { { b }^{ 2 } }{ c } \right) \)
\(\Rightarrow \left( -\frac { 12(1) }{ 4 } ,\frac { 4 }{ 4 } \right) \Rightarrow (-3,1)\)
∴The point of contact is (-3, 1).

Given equation of hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 64 } =1\)
Let y = mx + c be the required tangent
⇒ a² = 16 and b² = 64
The tangents are parallel to 10x - 3y + 9 = 0.
∴ Slope of tangent (m)
= Slope of the line 10x - 3y + 9 = 0.
∴ m = \(\frac{-co - efficient of \ x}{co - efficient of \ y}\)= \(\frac { -10 }{ -3 } =\frac { 10 }{ 3 } \)
The condition for y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
⇒ \({ c }^{ 2 }=16\left( \frac { 100 }{ 9 } \right) -64=\frac { 1600-64\times 9 }{ 9 } \)
⇒ \(\frac { 1024 }{ 9 } \Rightarrow c=\pm \frac { 32 }{ 3 } \)
∴ The required tangents are
y = \(\frac { 10 }{ 3 } x+\frac { 32 }{ 3 } \) or \(y=\frac { 10x }{ 3 } -\frac { 32 }{ 3 } \)
⇒ 3y = 10x + 32 or 3y = 10x - 32
⇒ 10x−3y+32 = 0, or 10x+3y−32 = 0

Let the general equation of the circle be
x² +y² +2gx + 2fy + c = 0 ........ (1)
It passes through points (1, 1), (2, -1) and(3, 2) .
Therefore,
2g+2f+c = -2 ........ (2)
4g−2f+c = -5 ........(3)
6g+4f+c = -13 ...... (4)
(2) – (3) gives −2g + 4f = 3 ... (5)
(4) – (3) gives 2g + 6f = -8  ....(6)
(5) + (6) gives f = \(-\frac { 1 }{ 2 } \) 
Substituting f = \(-\frac { 1 }{ 2 } \) in (6), g = -\(\frac { 5 }{ 2 } \)
Substituting f = \(-\frac { 1 }{ 2 } \) and g = - \(\frac { 5 }{ 2 } \) in (2), c = 4 .
Therefore the required equation of the circle is
x²+ y²+ 2\(\left( -\frac { 5 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) \)y+4 = 0 and x² + y² − 5x − y + 4 = 0.

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Let O₁ O₂ be the centres of the two semi circular vents.
First vent with centre O₁ (12, 0) and radius r = 10 yields equation to first semicircle as
(x−12)²+(y− 0)² = 10²
\(\Rightarrow\) x²+y²−24x + 44 = 0, y > 0
Second vent with centre O₂ (34, 0) and radius r = 10 yields equation to second vent as
(x−34)²+ y² = 10² 
x²+y²− 68x  + 1056 = 0, y > 0