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 line is y = x + c
m = 1, c = c
Equation of the hyperbola is 9x² - 16y² = 144
\(\div \)144 we get \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
a² = 16, b² = 9
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ c² = 16(1)² - 9 = 16 - 9 = 7
∴ c = ±\(\sqrt7\)

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\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

Given 2b = 5 and 2ae = 13
b² = a²( e² - 1) - b ⇒ a \(\sqrt { { e }^{ 2 }-1 } \)
2b = 5 ⇒ 2a\(\sqrt { { e }^{ 2 }-1 } \) = 5
⇒ 4a²( e² - 1) = 25 [squaring both sides]
⇒ 4a²e²- 4a² = 25
⇒ (2ae)² - 4a² = 25
⇒ 13²-4a²=25 [∵ 2ae=13]
⇒169- 25 = 4a²
⇒ 4a²= 144
⇒ a²= 36
⇒ a = 6
∴ 2b = 5 ⇒ b = \(\frac52\)⇒b² = \(\frac{25}{4}\)
∴ Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ 36 } -\frac { { y }^{ 2 } }{ \frac { 25 }{ 4 } } =1\)
\(\frac { { x }^{ 2 } }{ 36 } -\frac { { 4y }^{ 2 } }{ 25 } =1\)

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 \(\frac { { x }^{ 2 } }{ \frac { 81 }{ 4 } } +\frac { { y }^{ 2 } }{ \frac { 45 }{ 4 } } =1\)
⇒ \(\frac { 4{ x }^{ 2 } }{ 81 } +\frac { 4{ y }^{ 2 } }{ 45 } =1\)

Since foci are on the y-axis, equation of the ellipse is 
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)(a ∴ b²)
b² = a²(1 - e²) = a²\(\left( 1-\frac { 9 }{ 16 } \right) ={ a }^{ 2 }\left( \frac { 7 }{ 16 } \right) \)
⇒ \(b=\frac { a\sqrt { 7 } }{ 4 } \)
∴ Equation of the ellipse is \(\frac { { x }^{ 2 } }{ \frac { 7{ a }^{ 2 } }{ 16 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\)
\(\frac { 16(36) }{ 7{ a }^{ 2 } } +\frac { 16 }{ { a }^{ 2 } } =1\)
This passes through (6, 4)
∴ \(\frac { 16{ x }^{ 2 } }{ { 7 }a^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\)
⇒ \(\frac { 16 }{ { a }^{ 2 } } \left( \frac { 36 }{ 7 } +1 \right) =1\) ⇒ \(\frac { 36+7 }{ 7 } =\frac { { a }^{ 2 } }{ 16 } \)
⇒ \(\frac { 43 }{ 7 } \times 16={ a }^{ 2 }\Rightarrow { a }^{ 2 }=\frac { 43\times 16 }{ 7 } \)
Substituting \({ a }^{ 2 }=\frac { 43\times 16 }{ 7 } \) is (1) we get,
\(\frac { { x }^{ 2 } }{ 7\left( \frac { 43\times 16 }{ 7 } \right) } +\frac { { y }^{ 2 } }{ \left( \frac { 43\times 16 }{ 7 } \right) } =1\)
\(\frac { { x }^{ 2 } }{ 688 } +\frac { 7{ y }^{ 2 } }{ 688 } =1\)

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]

Given line in Ix + my + n = 0 ....(1)
tangent at (x₁, y₁) to the circle x² + y² = 9² is
xx₁ + yy₁= a₂ ...(2)
Comparing the co-efficients of like terms in (1)
and (2), we get, \(\frac { { x }_{ 1 } }{ l } =\frac { { y }_{ 1 } }{ m } =\frac { -{ { a }^{ 2 } } }{ n } \)
\({ x }_{ 1 }=\frac { -{ a }^{ 2 }l }{ n } \), and \({ y }_{ 1 }=\frac { -{ a }^{ 2 }m }{ n } \)
Since (x₁ , y₁) is a point on the circle, x²₁ + y²₁ = a²
\(\left( \frac { -{ a }^{ 2 }l }{ n } \right) +\left( \frac { -{ a }^{ 2 }m }{ n } \right) ={ a }^{ 2 }\)
\(\frac { -{ a }^{ 4 }{ l }^{ 2 } }{ { n }^{ 2 } } +\frac { { a }^{ 4 }{ m }^{ 2 } }{ { n }^{ 2 } } ={ a }^{ 2 }\)
⇒ \(-{ a }^{ 4 }{ l }^{ 2 }+{ a }^{ 4 }{ m }^{ 2 }={ a }^{ 2 }\)
\({ a }^{ 2 }({ l }^{ 2 }+{ m }^{ 2 })=1\)

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

Given equation is x² + y² - 2x + 5y + 7 = 0
Here 2g = -2 ⇒ g = -1 ⇒ 2f = 5 ⇒ f = \(\frac { 5 }{ 2 } \)
c = 7
Centre is (-g, -f) = \(\left( 1,\frac { -5 }{ 2 } \right) \)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { { 1 }^{ 2 }+{ \left( \frac { -5 }{ 2 } \right) }^{ 2 }-7 } \)
= \(\sqrt { 1+\frac { 25 }{ 4 } -7 } =\sqrt { \frac { 25 }{ 4 } -6 } =\sqrt { \frac { 1 }{ 4 } } =\frac { 1 }{ 2 } \)
∴ Cireumferenee of elrele = 2πr = 2π\(\left( \frac { 1 }{ 2 } \right) \) = π units
Area of the circle = πr² = π\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\) = \(\frac { \pi }{ 4 } \) sq.units