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 = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { { b }^{ 2 } }{ { 2a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { 1 }{ \sqrt { 2 } } \)

Given equation is x² + 3y² = a²
\(\div \) a² we get, \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ \frac { { a }^{ 2 } }{ 3 } } =1\)
Here a² and b2 = \(\frac{a^2}{3}\) ⇒ b = \(\frac{a}{\sqrt3}\)
Length of major axis is 2a and
Length of minor axis is 2b = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Taking (x₁, y₁) as (2, 3), we get
x₁² + y₁² + 2gx₁+ 2fy₁+ c = 2² + 3² − 6 × 2 − 8 × 3 + 12 
= 4 + 9 - 12 - 24 + 12
= -11\(<\)0.
Therefore the point (2, 3) lies inside the circle, by theorem.

Equation of the circle with end points of the diameter as (x₁, y₁₎ and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x² + y² + 3x + y − 6 = 0 which is the required equation of the circle.

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 centre is (2, -1) and passing through the point (3, 6)
∴ r = distance between (2, -1) and (3, 6)
= \(\sqrt { { (2-3) }^{ 2 }+(-1{ -6) }^{ 2 } } \)
= \(\sqrt { { (-1) }^{ 2 }+({ -7) }^{ 2 } } \)
= \(\sqrt { 1+49 } =\sqrt { 50 } \)
∴ Equation of the circle is
(x - h)² + (y - k)² = r²
(x−2)²+(y+1)² = \({ (\sqrt { 50 } })^{ 2 }\)
(x−2)²+(y+1)² = 50

Coefficient of x² = Coefficient of y² (characteristic (ii) for a second degree equation to represent a circle).
That is, 3 = a + 1 and a = 2 .
Therefore the equation of the circle is 
3x² + 3y² + 6x − 9y + 6 = 0
x² + y² + 2x − 3y + 2 = 0
So, centre is \(\left( -1,\frac { 3 }{ 2 } \right) \) and radius  r =\(\sqrt { 1+\frac { 9 }{ 4 } -2 } \)
=\(\sqrt { \frac { 5 }{ 2 } } \)

As the circle touches both the axes, the distance of the centre from both the axes is 3 units, centre can be (±3, ±3) and hence there are four circles with radius 3, and the required equations of the four circles are
x² + y² ± 6x ± 6y + 9 = 0.

9x²- y²- 36x - 6y + 18 = 0
Given equation is 9x²- y²- 36x - 6y + 18 = 0
⇒ 9x² - 36x - (y² + 6y) = -18
⇒ 9(x²-4x)-(y²+6y) =-18
⇒ 9(x² - 4x + 4 - 4) - (y² + 6y + 9 - 9) = -18
⇒ 9(x-2)²-36-(y+3)²+9 =-18
⇒ 9(x-2)² - (y+3)² = -18+36-9
⇒ 9(x - 2)² - (y + 3)² = 9
Dividing by 9 we get, \(\frac { { (x-2) }^{ 2 } }{ 1 } -\frac { ({ y+3) }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola whose transverse axis is parallel to x-axis.
a² = 1, b² = 9
∴ c² = a² + b² = 1 + 9 = 10 ⇒ c = \(\sqrt { 10 } \)
\(e=\sqrt {1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }} =\sqrt { 1-\frac { 9 }{ 1 } } =\sqrt { 10 } \)
a) Center is (2, -3)
⇒ h = 2, k = -3
(b) Foci are (h + c, k), (17 - c, k)
⇒ (2 +\(\sqrt { 10 } \), -3), (2 - \(\sqrt { 10 } \), -3)
(c) Vertic ar (h + a, k) (h - a, k)
⇒ (2+ 1,-3), (2-1,-3)
⇒ (3, -3) (1, -3)
(d) Equation of directrices are x - 2 = \(\pm \frac { a }{ e } \)
⇒ \(x-2=\pm \frac { 1 }{ \sqrt { 10 } } \)
\(x=2\pm \frac { 1 }{ \sqrt { 10 } } \)
⇒ \(x=2+\frac { 1 }{ \sqrt { 10 } } \) and \(x=2-\frac { 1 }{ \sqrt { 10 } } \)

Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { y^{ 2 } }{ 9 } =1\)
∴ a² = 25, b² = 9
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
Focus is (ae, 0) = \(\left( 5\times \frac { 4 }{ 5 } \right) \) = (4, 0)
Since the focus of the hyperbola coincides with the focus of the ellipse, foci of the hyperbola are (±4,0).
Let A be the length of the semi-transverse axis
∴ Ae - 4 ⇒ 2A = \(\frac { 4 }{ e } =\frac { 4 }{ 2 } =2\) [∵ e = 2]
Let B b th length of the semi conjugate axis
B² = A²(e² - 1) = 4(4 - 1) = 12
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

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).

(3) c² = a²(1+m²) 1, 2, 4 are related to parabola