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.

Here A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0, A = 0 either A or C is 0.
Hence, the given equation represents a parabola.

Here A = 3, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

Here A = 2, B = 0, C = -1, F = -7
Here A ≠ C and A and C are of opposite signs.
Hence the given equation represents a hyperbola.

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

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

Equation of the parabola is y² = 4ax
Latus rectum LL′ passes through the focus (a, 0)  
Hence the point L is (a, y₁) 
Therefore y₁² = 4a²
Hence y₁ = ±2a
The end points of latus rectum are (a, 2a) and (a, -2a)  

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.

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

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 \(x=2-\frac { 1 }{ \sqrt { 10 } } \)

18x²+ 12y² - 144x + 48y + 120 = 0
Given equation is
18x² + 12y² - 144x + 48y + 120 = 0
18x² - 144x + 12y² + 48y = -120
⇒ 18(x² - 8x) + 12(y² + 4y) = -120
⇒ 18(x²-8x+ 16-16)+ 12(y² +4y+4-4) =-120
18(x - 4)² - 288 + 12 (y + 2)²- 48 = -120
⇒ 18(x - 4)²+ 12(y + 2)² = -120 + 288 + 48
⇒ 18(x - 4)²+ 12(y + 2)² = 216
Dividing by 216 we get,
\(\frac { { 18(x-4) }^{ 2 } }{ 216 } +\frac { 12({ y+2) }^{ 2 } }{ 216 } =1\)
\(\Rightarrow \frac { { (x-4) }^{ 2 } }{ 12 } +\frac { ({ y+2) }^{ 2 } }{ 18 } =1\)
This is an equation of the ellipse with major axis parallel to y-axis,
∴ a² = 18, b² = 12
∴ c² = a² - b² = 18 -12 = 6 ⇒ c = \(\sqrt { 6 } \)
e =\( \sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 12 }{ 18 } } =\sqrt { \frac { 18-12 }{ 18 } } \)
\(=\sqrt{\frac{\not 6^1}{\not{18}}_{3}}=\sqrt{\frac{1}{3}}\)= \(\frac{1}{\sqrt 3}\)
(a) Center is (4, -2)
⇒ h = 4, k = -2
(b) Vertices are (h, k-a), (h, k + a)
⇒ (4, -2 - 3\(\sqrt { 2 } \)), (4, -2 + 3\(\sqrt { 2 } \))
[∴ a² = 18 ⇒ a = \(\sqrt { 18 } \) = 3\(\sqrt { 2 } \)]
(c) Foci are (h, k - c), (h, k + c)
⇒ (4, -2 - \(\sqrt { 6 } \)), (4, -2 + \(\sqrt { 6 } \))
(d) Equation of directrices are y + 2 = \(\pm \frac { a }{ e } \)
⇒ y+ 2 = \(\pm \frac { a }{ e } \)
\(\Rightarrow y-2=\pm \frac { 3\sqrt { 2 } }{ \frac { 1 }{ \sqrt { 3 } } } =\pm 3\sqrt { 2 } \times \sqrt { 3 } =\pm 3\sqrt { 6 } \)
\(\Rightarrow y+2=\pm 3\sqrt { 6 } ,y+2=-3\sqrt { 6 } \)
\(\Rightarrow y=-2+3\sqrt { 6 } \) and \( y=-2-3\sqrt { 6 } \)

\(\frac { { \left( x+3 \right) }^{ 2 } }{ 225 } -\frac { { \left( y-4 \right) }^{ 2 } }{ 64 } =1\)
Given equation is \(\frac { { \left( x+3 \right) }^{ 2 } }{ 225 } -\frac { { \left( y-4 \right) }^{ 2 } }{ 64 } =1\)
This is an equation of the hyperbola
∴ a² = 225, b² = 64
⇒ c² = a² + b² = 225 + 64 = 289
⇒ c = 17
\(e=​​​​\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 64 }{ 225 } } =\sqrt { \frac { 225+64 }{ 225 } } \)
= \(\sqrt { \frac { 289 }{ 225 } } =\frac { 17 }{ 15 } \)
(a) Center is (-3, 4)
⇒ h = -3, k = 4
(b) Foci are (h + c, k), (h - c, k)
⇒ (-3 + 17,4), (-3 -17, 4)
⇒ (14, 4) (-20, 4)
(c) Vertices are (h + a, k) and (h - a, k)
⇒ (-3 + 15,4), (-3 - 15,4)
⇒ (12, 4) (-18, 4)
(d) Equation of directrices are x + 3 = \(\pm \frac { a }{ e } \)
\(\Rightarrow x+3=\pm \frac { 15 }{ \frac { 17 }{ 15 } } \Rightarrow x+3=\pm \frac { -225 }{ 17 } \)
\(\Rightarrow x=\frac { 225 }{ 17 } \) and \(x=\frac { -225 }{ 17 } \)
\(\Rightarrow x=\frac { 225-51 }{ 17 } \) and \(x=\frac { -225-51 }{ 17 } \)
\(\Rightarrow x=\frac { 174 }{ 17 } \) and \(x=\frac { -276 }{ 17 } \)

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 144 } =1\)

This is an equation of the hyperbola.
∴ a² = 25 and b² = 144
⇒ c² =a² + b² =25 + 144 =169 ⇒ c = 13
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 144 }{ 25 } } =\sqrt { \frac { 169 }{ 25 } } =\frac { 13 }{ 5 } \)
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) Foci are (h + c, k), (h - c, k)
⇒ (0 + 13,0), (0 - 13,0)
⇒ (13, 0), (-13, 0)
(c) Vertices are (h + a, k) and (h - a, k)
⇒ (0 + 5, 0), (0 - 5, 0) ⇒ (5, 0), (-5, 0)
(d) Equations of Directrices are x = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow x=\frac { 5 }{ \frac { 13 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 13 } \)

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

As per the given information, we can take the parabola as open downward.

∴ Its equation is x² = -4ay ...(1)
Let P be a point on the flow paths, 2.5 m below the line of the pipe and 3m beyond the vertical line through the end of the pipe.

∴ Pis (3, -2.5)
∴ (1) becomes 3² = -4a(-2.5)
⇒  \(\frac { -9 }{ 2.5 } \) = 4a
∴ (1) becomes, x₁² = \(\frac { -9 }{ 2.5 } \)y .....(2)
Let x₁ be the distance between the bottom of the verticallineon the ground from the pipeendand the point on which this water touches the ground.
But the height of the pipe from the ground is 7.5 m.
∴ (x₁, -7.5) lies on ....(2)
⇒ x₁² = \(\frac { -9 }{ 2.5 } \)(-7.5)
⇒ x₁ = \(\sqrt { 9\times 3 } =3\sqrt { 3 } \) m
The water strikes the ground \(3\sqrt { 3 } \) m beyond the vertical line.

Let AB be the rod and P(x₁, y₁) be a point on the rod such that AP = 0.3 m.
Draw PD ⊥ x-axis and PC ⊥ y - axis.

Δ ADP ≅ Δ PCB
∴ \(\frac { PC }{ DA } =\frac { PB }{ AP } =\frac { BC }{ PD } \)
⇒ \(\frac { x_{ 1 } }{ DA } =\frac { 0.9 }{ 0.3 } =\frac { BC }{ { y }_{ 1 } } \)
⇒ \(DA=\frac { 0.3{ x }_{ 1 } }{ 0.9 } =\frac { { x }_{ 1 } }{ 3 } \)
and BC = \(\frac { 0.9{ y }_{ 1 } }{ 0.3 } =\frac { 9 }{ 3 } { y }_{ 1 }=3{ y }_{ 1 }\)
Now OA = OD + DA
= \({ x }_{ 1 }+\frac { { x }_{ 1 } }{ 3 } =\frac { 4{ x }_{ 1 } }{ 3 } \)
OB = OC + BC = y₁ + 3y₁ = 4y₁
But OA² + OB² = AB²
⇒ \({ \left( \frac { 4{ x }_{ 1 } }{ 3 } \right) }^{ 2 }+{ \left( 4{ y }_{ 1 } \right) }^{ 2 }={ \left( 1.2 \right) }^{ 2 }\)
⇒ \(\frac { { { x }_{ 1 } }^{ 2 } }{ 9 } +\frac { { { y }_{ 1 } }^{ 2 } }{ 9 } =\frac { 1.44 }{ 16 } =0.09\) ≅ 1
∴ Locus of (x₁, y₁) is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 1 } =1\)
Here a² = 9, b² = 1
∴ \(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 9 } } =\sqrt { \frac { 9-1 }{ 9 } } \)
= \(\sqrt { \frac { 8 }{ 9 } } \)
e = \(\frac { 2\sqrt { 2 } }{ 3 } \) 

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.

The parametric equation of tangent at 't₁' to the parabola y² = 4ax is yt₁ = x + at₁²  ...(1)
Also, the parametric equation of tangent at 't₂' to the parabola y = 4ax is yt₂ = x+ at₂² ...(2)
(1) ➝ yt₁ = x + at₁²
(2) ➝ yt₂ = x + at₂²
(1) - (2) y(t₁ -t₂) = a(t₁² - t₂²)

⇒ y = a(t₁ + t₂)
Substitutingy = a(t₁ + t₂) in (1) we get,
a(t₁ + t₂)t₁ = x + at₁²

⇒ x = at₁t₂
Hence, the point of intersection of two lengths is
[at₁t₂, a(t₁ + t₂)]

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the ellipse.
a² = 2⁵ and b² = 9 and c² = a² - b²
⇒ c² = 25 - 9 = 16 ⇒ c = 4
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) foci are (h - c, k), (h + c, k)
⇒ (0 - 4, 0), (0 + 4, 0)
⇒ (-4, 0) and (4, 0)
(c) Vertices are (h - a, k) and (h + a, k)
⇒ (0 - 5, 0) and (0 + 5, 0)
⇒ (-5, 0) and (5, 0)
(d) Directrices are x = \(\pm \frac { a }{ e } \)
⇒ x = \(\pm \frac { 5 }{ e } \)
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
∴ Directrice are x = \(\pm \frac { 5 }{ \frac { 4 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 4 } \)