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

From figure the midpoint of line joining foci is the centre C(0, 0).
Transverse axis is y-axis 
AA′ = 2a \(\Rightarrow \) 2a = 8,
SS′ = 2c = 12, c = 6
a = 4
b² = c²−a² = 36−16 = 20
Hence the equation of the required hyperbola is \(\frac { { y }^{ 2 } }{ 16 }- \frac { { x }^{ 2 } }{ 20 } =1\)

Since the circles touch both the axes. Its equation will be
(x + a)² + (y + a)² = a² ...............(1)
It passes through (-4, -2)
∴ (-4 + a)² + (-2 + a)² = a²
\(16+\not a^{2}+8 a+4+a^{2}+4 a=\not a^{2}\)
⇒ a² + 12a + 20 = 0
⇒ (a + 10)(a + 2) = 0
a = -10 or -2

Case (i):
When a = -10, (1) becomes
(x + 10)² + (y + 10)2 = 10²
\(\Rightarrow x^{2}+\not 100+20 x+y^{2}+\not 100+20 y=160\)
⇒  x² + y²+ 20x + 20y + 100 = 0
Case (ii):
When a = -2, (1) becomes
⇒ (x + 2)² + (y + 2)² = 2²
\(x^{2}+4 x+4+y^{2}+4 y+\not 4 = \not 4\)
x² + y²+ 4x + 4y + 4 = 0
Hence, equation of the circles are
x² + y²+ 4x + 4y + 4 = 0
or  x² + y²+ 20x + 20y + 100 = 0

Given r = 5 cm

Since the circle touches the x axis, its centre is (0, ±5)
Equation of the circle is (x - h)² + (y - k)² = r²
⇒ (x - 0)² + (y ± 5)² = 25
\(\Rightarrow x^{2}+y^{2}+\not 25 \pm 10 y=\not 25\)
⇒ x²+y²+10y = 0

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

Equation of tangent to the circle at P(x₁, y₁ ) is xx₁ yy₁ = a²
That is, x(−3) + y(4) = 25
−3x + 4y = 25
Equation of normal is xy₁ - yx₁ = 0
That is, 4x + 3y = 0 

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 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 \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Given ends of diameter are (3, 4)(2, -7)
∴ Equation of the circle is
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
⇒ (x - 3)(x - 2) + (y - 4)(y + 7) = 0
⇒ x² - 2x - 3x + 6 + y² + 7y - 4y - 28 = 0
⇒ x² + y² − 5x + 3y − 22 = 0

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 equation of the parabola be y² = 4ax
Let AB = I.

Since ABC is an equilateral triangle,
∠BAM = ∠MAC = 30^o
In MBM, cos 30°= \(\frac{AM}{l}\)
⇒ AM = l cos 30° = l\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \)
sin 30^o = \(\frac { BM }{ AM } \)
⇒ sin 30^o = \(\frac { BM }{ l } \)
⇒ BM = l sin 30^o = l\(\left( \frac { 1 }{ 2 } \right) \)
ஃ Co-ordinates of B are (AM, BM) ⇒ \(B\left( \frac { \sqrt { 3 } l }{ 2 } ,\frac { 1 }{ 2 } \right) \)
Also B lies on (1)
\({ \left( \frac { l }{ 2 } \right) }^{ 2 }=4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
⇒ \(\frac { { l }^{ 2 } }{ 4 } =4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
\(\frac { { l }^{ 2 } }{ 4 } =2\sqrt { 3 } al\)
\(\frac { l }{ 4 } =2\sqrt { 3 } \) a ⇒ 1 = 8 \(a\sqrt { 3 } \)
Length of the side of the equilateral triangle is 8 \(a\sqrt { 3 } \) units.

Equation of ellipse is
x²+ 4y² = 32
\(\frac { { x }^{ 2 } }{ 32 } + \frac { { y }^{ 2 } }{ 8 } =1\)
a² = 32, b² = 8
\(a=4\sqrt { 2 } ,b=2\sqrt { 2 } \)
Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is \(\frac { xcos\frac { \pi }{ 4 } }{ 4\sqrt { 2 } } \frac { ysin\frac { \pi }{ 4 } }{ 2\sqrt { 2 } } =1\) 
\(\frac { x }{ 8 } +\frac { y }{ 4 } =1\)
x+2y−8 = 0.
Equation of normal is \(\frac { 4\sqrt { 2X } }{ cos\frac { \pi }{ 4 } } -\frac { 2\sqrt { 2Y } }{ sin\frac { \pi }{ 4 } } =32-8\)
That is 8x-4y = 24 
2x-y-6 = 0
Aliter: 
At, \(\theta =\frac { \pi }{ 4 } \)
\((a\ cos \theta ,b\ sin \theta )=\left( 4\sqrt { 2 }\ cos\frac { \pi }{ 4 } ,2\sqrt { 2 }\ sin\frac { \pi }{ 4 } \right) \)
= (4, 2)
∴ Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is same at (4, 2) 
Equation of tangent in cartesian form is \(\frac { { xx }_{ 1 } }{ { a }^{ 2 } } +\frac { { yy }_{ 1 } }{ { b }^{ 2 } } =1\)
x+2y−8 = 0
Slope of tangent is -\(\frac { 1 }{ 2 } \)
Slope of normal is 2 Equation of normal is  y - 2 = 2(x -4) 
y−2x+6 = 0 

Rearranging terms in the equation of hyperbola to bring it to standard form,
we have, 11(x²-4x)-25(y²-2y)-256 = 0
11(x− 2)²−25(y−1)² = 256−44+25
11(x−2 )²− 25 (y−1)² = 275
\(\frac { { \left( x-2 \right) }^{ 2 } }{ 25 } -\frac { { \left( y-1 \right) }^{ 2 } }{ 11 } =1\)
Centre (2, 1) a² = 25, b² = 11
c² = a² +b²
= 25 +11 = 36
Therefore, c = ±6
and e = \(\frac { c }{ a } =\frac { 6 }{ 5 } \)and the coordinates of foci are(8, 1) and(-4, 1) from figure.

By the definition of a conic  \(\frac{SP}{PM}\)= e or SP² = e²PM²
Then, (x−2)² + (y−3)² = \(\frac { 1 }{ 4 } \) (x-7)² 
3x²+ 4y²−2x − 24y + 3 = 0
\({ 3\left( x-\frac { 1 }{ 3 } \right) }^{ 2 }+4(y-3)^{ 2 }=3\left( \frac { 1 }{ 9 } \right) +4\times 9-3=\frac { 100 }{ 3 } \)
\(\frac { { \left( x-\frac { 1 }{ 3 } \right) }^{ 2 } }{ \frac { 100 }{ 9 } } +\frac { (y-3{ ) }^{ 2 } }{ \frac { 100 }{ 12 } } \) = 1 which is in the standard form.
Therefore, the length of major axis = 2a = 2\(\sqrt { \frac { 100 }{ 9 } = } \frac { 20 }{ 3 } \) and
the length of minor axis = 2b = 2\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).