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.

Since the parabola is symmetric about x-axis, it is either open upward or downward.
Let the equation be x² = 4ay ...(1)
Since (2, -3) lies on the parabola,
2² = 4a(-3) ⇒ a = \(\frac { -1 }{ 3 } \)
Substituting a = \(\frac { -1 }{ 3 } \) in (1) we get,
x² = 4 \(\left( \frac { -1 }{ 3 } \right) \) y ⇒ 3x² = -4y. Which is the required equation of the parabola.

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 + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

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 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 vertex be (0, 0) .
The equation of the parabola is
y² = 4ax
(1) Since the diameter is 40cm and the depth is 30 cm , the point (30, 20) lies on the parabola.
20² = 4a × 30
4a = \(\frac { 400 }{ 30 } \) = \(\frac { 40 }{ 3 } \)
Equation is y² = \(\frac { 40 }{ 3 } \)x.
(2) The bulb is at focus (0, a).
Hence the bulb is at a distance of \(\frac { 10 }{ 3 } \)cm from the vertex.

From the graph the vertex is at (0, 0) and the parabola is open down
Equation of the parabola is x² = -4ay
(-20, -15) and (20, -15) lie on the parabola 
20² = -4a(-15)
\(4a=\frac { 400 }{ 15 } \)
x² =\(\frac { -80 }{ 3 } \) x y
Therefore equation is 3x² = -80y

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

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]

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, -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

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 

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 = \(\frac{2\sqrt5}{2}\) = \(\sqrt5\)

The cross section of a nuclear cooling tower is in the shape of a hyperbola.

GIven OC = \(\frac12\) OD and CD = 150 m
Its equation is OC = 50 m & OD = 100 m
\(\frac { { x }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { y }^{ 2 } }{ { 44 }^{ 2 } } =1\)  ......(1)
Let I be the radius of the top of the tower
∴ A(l, 50) is a point on the hyperbola
∴ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } \frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =1\)
⇒ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 50 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ l }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 2500)
⇒ l² = \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \)(66.60)
⇒ \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \) = 45.41 m
Radius of the top of the tower is 45.41 m.
Let h be the radius of the base of the tower.
∴ B(h, 100) is a point on the hyperbola
∴ (1) becomes
\(\frac { { h }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =1\Rightarrow \frac { h^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 100 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ h }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 10000)
⇒ \({ h }^{ 2 }=\frac { 30 }{ 44 } \sqrt { 11936 } =\frac { 30 }{ 44 } \)(109.25)
⇒ \(h=\frac { 3277.5 }{ 44 } \) = 74.48 m.
Radius of the base of the tower is 74.48 m.
Diameter of the base = 148.96 m
Diameter of the top and base of the tower are 90.82 m and 148.96 m.

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.

Completing the square on x and y of 4x²+36y²+40x−288y+532 = 0,
4(x² + 10x + 25 − 25) + 36(y² − 8y + 16 − 16) + 532 = 0 , gives
4(x² + 10x + 25) + 36(y² − 8y + 16) = −532 + 100 + 576
4(x + 5)² + 36(y − 4)² = 144.
Dividing both sides by 144, the equation reduces to \(\frac { { \left( x+5 \right) }^{ 2 } }{ 36 } \frac { { \left( y-4 \right) }^{ 2 } }{ 4 } =1\)
This is an ellipse with centre (-5, 4), major axis is parallel to x-axis, length of major axis is 12 and length of minor axis is 4. Vertices are (1, 4) and (-11, 4).
Now, c² = a²−b² = 36 − 4 = 32
and c = ±4 \(\sqrt { 2 } \)
Then the foci are (−5 − 4\(\sqrt { 2 } \), 4) and (−5 + 4\(\sqrt { 2 } \) , 4).
Length of the major axis = 2a = 12 units and
the length of the minor axis = 2b = 4 units.

Let the general equation of the circle be
x² +y² +2gx + 2fy + c = 0 ........ (1)
It passes through points (1, 1), (2, -1) and(3, 2) .
Therefore,
2g+2f+c = -2 ........ (2)
4g−2f+c = -5 ........(3)
6g+4f+c = -13 ...... (4)
(2) – (3) gives −2g + 4f = 3 ... (5)
(4) – (3) gives 2g + 6f = -8  ....(6)
(5) + (6) gives f = \(-\frac { 1 }{ 2 } \) 
Substituting f = \(-\frac { 1 }{ 2 } \) in (6), g = -\(\frac { 5 }{ 2 } \)
Substituting f = \(-\frac { 1 }{ 2 } \) and g = - \(\frac { 5 }{ 2 } \) in (2), c = 4 .
Therefore the required equation of the circle is
x²+ y²+ 2\(\left( -\frac { 5 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) \)y+4 = 0 and x² + y² − 5x − y + 4 = 0.