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 \(\frac { { y }^{ 2 } }{ 49 } -\frac { { x }^{ 2 } }{ \frac { 343 }{ 9 } } =1\) 
⇒ \(\frac { { y }^{ 2 } }{ 49 } -\frac { 9{ x }^{ 2 } }{ 343 } =1\)

Since the foci are one the x-axis, the equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given conjugate axis = \(\frac34\) (transverse axis)
⇒ 2b = \(\frac34\) (2a) ⇒ b = \(\frac{3a}4\) ⇒ b² = \(\frac { { 9a }^{ 2 } }{ 16 } \)
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9{ a }^{ 2 } }{ { 16a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
∴ e = \(\frac { 5 }{ 4 } \)

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 = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Given equation of tangent is y = 3x + 1
The condition for any line y = mx + c to be a tangent to y² = 4ax is c = \(\frac{a}{m}\)
1= \(\frac{a}{m}\) ⇒ 1 = \(\frac{a}{m}\) ⇒ a = 3
Length of the latus rectum is 4a = 4(3) = 12 units.

Given circle is  x² +y² + 2x - 3y - 8 = 0
Equation of tangent is xx₁ + yy₁ + 1(x + x₁) -\(\frac{3}{2}\)
(y + y₁) - 8 = 0
AE(2, 3), the tangent is
x(2) + y(3) + x + 2 - \(\frac{3}{2}\) (y + 3) -8 = 0
⇒ 3x + 3y + 2 - \(\frac{3y}{2}\) - \(\frac92\) - 8 = 0
Multiply by 2 we get,
⇒ 6x + 6y + 4 - 3y - 9 - 16 = 0
⇒ 6x + 3y - 21 = 0

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Equation of the circle is x² + (y + 2)² = 0
Compare with(x-h)²+(y-k)² = r²
h = 0, k = -2, r² = 0
Centre (h, k) = (0, -2)
radius is 0.

Equation of the circle in standard form is (xr − h)² + (y − k)² = r²
\( \Rightarrow (x-(-3))^{2}+(y-(-4))^{2} =3^{2} \)
\( \Rightarrow (x+3)^{2}+(y+4)^{2} =3^{2} \)
\( \Rightarrow x^{2}+y^{2}+6 x+8 y+16 =0 .\)

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

y² = -8x
The given parabola is left open parabola
and 4a = 8 ⇒ a = 2
(a) Vertex is (0, 0)
⇒ h = 0, k = 0
(b) focus is (h - a, 0 + k)
⇒ (0-2, 0 + 0)
⇒ (-2, 0)
(c) Equation of directrix is x = h + a
⇒ x = 0 + 2 ⇒ x = 2
(d) Length of latus rectum is 4a = 8.

Length of latus rectum = 4,
distance between foci = 4\(\sqrt { 2 } \) major axis is y-axis
Given \(\frac { { 2b }^{ 2 } }{ a } =4\) and distance between foci = 2ae = 4\(\sqrt { 2 } \)
⇒ ae = \(2\sqrt { 2 } \)
⇒a²e² = 8 ....(1)
\(\frac { { 2b }^{ 2 } }{ a } =4\Rightarrow { b }^{ 2 }=2a\) ...(2)
We know b² = a²(1 - e²)
b² = a² - a²e²
2a = a2 - 8
[using (1) and (2)]
a² - 2a - 8 = 0
On factorising we get
(a - 4)(a + 2) = 0
a = 4 or-2
a = 4
⇒ [∴ a = -2 is not possible]
⇒ ae = 16
∴ From (2), b² = 2(4) = 8
Hence, the equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 8 } +\frac { { y }^{ 2 } }{ 16 } =1\) [∵ Major axis is y -axis]

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 \(\frac { { y }^{ 2 } }{ 16 }- \frac { { x }^{ 2 } }{ 20 } =1\)

Given equation of the circle is
3x² + (3 - p)xy + qy² - 2px = 8pq
For the circle, co-efficient of xy = 0
⇒ 3-p = 0 ⇒ p = 3
Also, co-efficient of x² = co-efficient of y²
⇒ 3 = q
∴ Equation of the circle is
3x² + 3y² - 6x = 8(3)(3)
3x² + 3y² - 6x - 72 = 0
Dividing by 3, we get
x² + y² - 2x - 24 = 0
Here 2g = - 2⇒ g = - 1
f = 0 and c = - 24
Centre is (-g, -f) (1, 0)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { { (-1 })^{ 2 }+0+24 } \)
= \(\sqrt { 25 } \) = 5 units.

Area of the circle = 9π sq.units

πr² = 9π ⇒ r² ⇒ 9 ⇒ r ⇒ 3
Diameters are x + y = 5 ................(1)
and x - y = 1 .................(2)
We know that centre is the point of intersection of diameters.
∴ To find the centre, solve (1) and (2).

⇒ 2x = 6
⇒ x = 3
∴ (1) ⇒ 3 + y = 5
⇒ y = 5 - 3 = 2
∴ Centre is (3, 2)
Hence, equation of the circle is
(x - h)² + (y - k)² = r2
⇒ (x - 3)² + (y - 2)² = 32
\(\Rightarrow x^{2}-6 x+9+y^{2}-4 y+4=9\)
⇒ x² + y² − 6x − 4y + 4 = 0

Given centre is (2, 3)
Let us solve 3x- 2y = 1  .....(1)
and 4x+ y = 27 .....(2)

⇒ x = 5
∴ 3(5) - 2y = 1
⇒ 15 - 2y = 1
⇒ 15-1 = 1
⇒ 14 = 2y
⇒ y = 7
 The circle passes through (5, 7)
[∵ distance between (5, 7) and (2, 3)]
r = \(\sqrt { { (5-2) }^{ 2 }+({ 7-3) }^{ 2 } } \)
= \(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \)
= \(\sqrt { 9+16 } =\sqrt { 25 } =5\)
Equation of the circle is
(x-h)²+(y-k)² = r²
(x - 2)² + (y - 3)² = 5²
x² - 4x + 4 + y² - 6y + 9 = 25
x² + y² - 4x - 6y + 13 - 25 = 0
x²+y²− 4x − 6y −12 = 0

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

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 

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 \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

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 =\(\sqrt { 1+\frac { 9 }{ 4 } -2 } \)
=\(\sqrt { \frac { 5 }{ 2 } } \)

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

Given F₁P + F₂P = 8
By the focal property of ellipse
F₁P + F₂P = 2a
∴ 2a = 8 ⇒ a = 4
and distance between the foci = F₁F₂ = 6
2ae = 6 ⇒ ae = 3
∴ 4(e) = 3 ⇒ e \(\frac34\)

∴ b² = a²(1- e²)
= \(16\left( { 1-\left( \frac { 3 }{ 4 } \right) }^{ 2 } \right) =16\left( 1-\frac { 9 }{ 10 } \right) =16\left( \frac { 7 }{ 16 } \right) =7\)
∴ The path traced by him is an ellipse and its equation is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 7 } \) = 1

Since the axis of the parabola is vertical, it is open upward.

Its equation is x² = 4ay .....(1)
Since the base of the parabola is 5 m
VA = 2.5 m and the height of the arch is 10 m
A(2.5, 10) lies on the parabola
(2.5)² = 4a(10)
⇒ 6.25 = 40a ⇒ a = \(\frac{625}{4000}=\frac{5}{32}\)
Substituting a = \(\frac{5}{32}\) in (1) we get,
x² = 4\(\left( \frac { 5 }{ 32 } \right) \) y ⇒ x² = \(\frac58\)y
Let x be the width of the arch, when the height is 2m.
∴ (x, 2) is a point on the parabola
∴ x² = \(\frac58\)(2) = \(\frac{5}{4}\)
∴ x = \(\frac{\sqrt5}{2}\)
and 2x = \(\frac{2\sqrt5}{2}\) = \(\sqrt5\)

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.

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

Equation of the ellipse is 2x² + 7y² = 14
\(\div \) we get, \(\frac { { x }^{ 2 } }{ 7 } +\frac { { y }^{ 2 } }{ 2 } =1\)
∴ a² = 7, b² = 2
The condition for the line y = mx + c to be a tangent to the ellipse is
\(y=mx\pm \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } } \) ....(1)
(5, 2) lies on (1) and a² = 7, b² = 2
∴ 2 = \(m(5)\pm \sqrt { 7{ m }^{ 2 }+2 } \)
2 - 5m = \(\pm \sqrt { 7{ m }^{ 2 }+2 } \)

Squaring both sides we get,
(2, - 5m)² = 7m² + 2
4 + 25m²- 20m = 7m² + 2
18m²-2m + 1 = 0 (\(\div \)2)
On factorising we get,
(m-1)(9m-1) = 0
∴ m -1 = 0 or
9m-1 = 0
⇒ m = 1 or \(\frac { 1 }{ 9 } \)
When m = 1, (1) becomes
y = \(1(x)\pm \sqrt { 7{ m }^{ 2 }+2 } \)
⇒ y = x \(x\pm \sqrt { 7+2 } \) ⇒ y = \(\pm \)3
y= x + 3 ⇒ x - y - 3 = 0
When m = \(\frac { 1 }{ 9 } \) (1) becomes
\(y=\frac { 1 }{ 9 } (x)\pm \sqrt { 7\left( \frac { 1 }{ 81 } \right) +2 } \)
⇒ \(y=\frac { x }{ 9 } +\sqrt { \frac { 7+162 }{ 81 } } \)
\(y=\frac { x }{ 9 } +\frac { 13 }{ 9 } \)
⇒ 9y = x + 13 ⇒ x -9y + 13 = 0
Hence the equation of tangents are x - y - 3 = 0 and x −9y + 13 = 0

Rearranging the terms, the equation of ellipse is 4x² + 24x + y²− 2y + 21 = 0

That is, 4(x² + 6x + 9 − 9) + (y² − 2y + 1 − 1) + 21 = 0,
 4(x + 3)² − 36 + (y−1)² −1 + 21 = 0,
 4(x + 3)² + (y − 1)² = 16,
\(\frac { { \left( x+3 \right) }^{ 2 } }{ 4 } +\frac { { \left( y+1 \right) }^{ 2 } }{ 16 } =1\)
Centre is (-3, 1) a = 4, b = 2, and the major axis is parallel to y-axis c² = 16−4 = 12
c = ±2\(\sqrt { 3 } \)
Therefore, the foci are (−3, 2\(\sqrt { 3 } \) +1) and (−3, −2\(\sqrt { 3 } \) +1).
Vertices are (3, ±4 +1).
That is the vertices are (3, 5) and (3, -3) and the length of Latus rectum = \(\frac { { 2b }^{ 2 } }{ a } \) = 2 units.

The Latus rectum LL′ of an ellipse \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) passes through S(ae, 0)  
Hence L is (ae, y₁ )
Therefore, \(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } +\frac { { y_1 }^{ 2 } }{ { b }^{ 2 } } \) = 1
\(\frac { { y _1}^{ 2 } }{ { b }^{ 2 } } \) = 1-e²
y₁²  = b²(1-e²)
= b²\(\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \)\(\left( since,{ e }^{ 2 }=1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \ \)
y₁ = \(\pm\frac { { b }^{ 2 } }{ { a } } \)
That is the end points of Latus rectum L and L′ are \(\left( ae,\frac { { b }^{ 2 } }{ { a } } \right) and\ \left( ae-\frac { { b }^{ 2 } }{ { a } } \right) \)
Hence the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ { a } } \)