Given line is x + y + 1 = 0
⇒ y = -x-1
m = -1, c = -1
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
a² = 16, b² = 15
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ (-1)² = 16(-1)² - 15
1 = 16 -15
1 = 1
Since the condition is satisfied, x + y + 1 = 0 touches the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
The point of contact is \(\left( \frac { -{ a }^{ 2 }m }{ c } ,\frac { -{ b }^{ 2 } }{ c } \right) \) = \(\left( \frac { -16(-1) }{ -1 } ,\frac { -15 }{ -1 } \right) \) = (-16, 15)
Hence, the point of contact is (-16, 15)

Given equation of the hyperbola is
3x² - 6y² = -18; \(\div \)by (-18) we get \(\frac{y^2}{3}- \frac{x^2}{6}\) = 1
The transverse axis is long y-axis.
Here a² = 3, b² = 6
Length of transverse axis is 2a = 2\(\sqrt { 3 } \)
Length of conjugate axis is 2b = 2\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

Let the equation of the ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given e = \(\frac { 2 }{ 3 } \) and \(\frac { 2{ b }^{ 2 } }{ a } \) = 5 ⇒ 2b² = 5a  ...(1)
∴ b² = a²(1 - e²) = a² \({ a }^{ 2 }\left( 1-\frac { 4 }{ 9 } \right) ={ a }^{ 2 }\left( \frac { 5 }{ 9 } \right) \)
∴ \({ 2b }^{ 2 }=\frac { 10{ a }^{ 2 } }{ 9 } \)  ...(2)
From (1) and (2),
\(\frac { 10{ a }^{ 2 } }{ 9 } \) = 5a ⇒ 10a² = 45a
10a² - 45a = 0 ⇒ 5a(2 - 9a) = 0
⇒ a = 0 or a = \(\frac { 9 }{ 2 } \) [∵ a = 0 is not possible]
∴ \({ a }^{ 2 }=\frac { 81 }{ 4 } \)
∴ \(2{ b }^{ 2 }=\frac { 5\times 9 }{ 2 } =\frac { 45 }{ 2 } \Rightarrow { b }^{ 2 }=\frac { 45 }{ 4 } \)
∴ Equation of the ellipse is \(\frac { { x }^{ 2 } }{ \frac { 81 }{ 4 } } +\frac { { y }^{ 2 } }{ \frac { 45 }{ 4 } } =1\)
⇒ \(\frac { 4{ x }^{ 2 } }{ 81 } +\frac { 4{ y }^{ 2 } }{ 45 } =1\)

Given equation is x² = -36y
focus is (0, -1) = (0, -9) and latus rectum is y = -9 in (1)

We get, x² = -36(-9) ⇒ x = \(\sqrt { 36(9) } \) = 6(3) = 18
Area of Δ AOB = 2(area of ΔOSB)
\(=2\left( \frac { 1 }{ 2 } \times b\times h \right) =2\left( \frac { 1 }{ 2 } \times 18\times 9 \right) =162\)
[∵ b = SB = 18, h = OS = 9]

Equation of circle is x² + y² = 64
∴ a² = 64 ⇒ a = 8
Given line is 3x+ 4y = P
4y = -3x + p
y = \(y=\frac { -3 }{ 4 } x+\frac { p }{ 4 } \)
m = \(\frac { -3 }{ 4 } \) and c = \(\frac { p }{ 4 } \)
The condition for y = mx + c to be a tangent to the circle in c² = a²(1 + m²).
∴ \({ \left( \frac { p }{ 4 } \right) }^{ 2 }=64\left( 1+\frac { 9 }{ 16 } \right) \)
\(\Rightarrow \frac{p^{2}}{\not 16}=64\left(\frac{16+9}{\not 16}\right) \Rightarrow p^{2}=64(25)\)
\(p=\pm \sqrt { 64(25) } =\pm 8(5)\)
∴ p = ±40