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 \( y=-2-3\sqrt { 6 } \)

\(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Given equatlon is \(\frac { { (y-2) }^{ 2 } }{ 25 } -\frac { ({ x+1) }^{ 2 } }{ 16 } =1\)
Thi is an equation of the hyperbola where transverse axis is parallel to y-axis.
∴ a² = 25, b² = 16
⇒ c² - a² + b² = 25 +16 = 41
⇒ c = \(\sqrt { 41 } \)
\(e =\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 16 }{ 25 } } =\sqrt { \frac { 41 }{ 25 } } =\frac { \sqrt { 41 } }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h, k + c), (h, k- c)
⇒ (-1, 2 + \(\sqrt { 41 } \)), (-1, 2 - \(\sqrt { 41 } \))
(c) Vertices are (h, k + a), (h, k-a)
⇒ (-1, 2 + 5), (-1, 2 - 5)
= (-1, 7), (-1, -3)
(d) Equation of directrices are
\(y-2=\pm \frac { 5 }{ \frac { \sqrt { 41 } }{ 5 } } \)
\(\Rightarrow y-2=\pm \frac { 25 }{ \sqrt { 41 } } \)
\(\Rightarrow y=2+\frac { 25 }{ \sqrt { 41 } } \) and 
\( y=2-\frac { 25 }{ \sqrt { 41 } } \)

\(\frac { { \left( x+3 \right) }^{ 2 } }{ 225 } -\frac { { \left( y-4 \right) }^{ 2 } }{ 64 } =1\)
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 \(x=\frac { -276 }{ 17 } \)

Given equation is \(\frac { { (x-1) }^{ 2 } }{ 100 } +\frac { ({ y-2) }^{ 2 } }{ 64 } =1\)
This is an equation of the ellipse
∴ a² = 100,b² = 64
⇒ c² = a² - b²
⇒ c² = 100 - 64 = 36
∴ c = 6
\(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 64 }{ 100 } } =\sqrt { \frac { 100-64 }{ 100 } } \)
\(=\sqrt{\frac{36}{100}}=\frac{\not 6} {\not {10}}\)
∴ c = \(\frac { 3 }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h - c, k), (h + c, k) ⇒ (-1 - 6, 2), (-1 + 6, -2) ⇒ (-7, 2), (5, 2)
(c) Vertices are (h - a, k) and (h + a, k) ⇒ (-1 -10,2), (-1 + 10,2) ⇒ (-11, 2), (9, 2)
(d) Equation of directrices are x + 1 = \(\pm \frac { a }{ e } \)
\(\Rightarrow x+1=\pm \frac { 10 }{ \frac { 3 }{ 5 } } \Rightarrow x+1=\pm \frac { 50 }{ 3 } \)
\(\therefore x+1=\frac { 50 }{ 3 } and x+1=-\frac { 50 }{ 3 } \)
\(\Rightarrow x=\frac { 50 }{ 3 } -1\) and \(x=-\frac { 50 }{ 3 } -1\)
\(\Rightarrow x=\frac { 47 }{ 3 } \) and \(x=\frac { -53 }{ 3 } \)

  \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Given equation is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola where the transverse axis is parallel to the y-axis.
∴ a² = 16, b² = 9, c² = a² + b²
⇒ c² = 16 + 9 = 25 ⇒ c = 5
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 16+9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
(a) Center is (0, 0)
⇒ h = 0, k = 0

(b) Vertices are (h, k + a), (h, k-a)
⇒ (0, 0 + 4), (0, 0 - 4) ⇒ (0, 4) (0,-4)
(c) Foci are (h, k + c), (h, k- c)
⇒ (0, 0 + 5), (0,  0 - 5) ⇒ (0, 5) (0, -5)
(d) Equations of Directrices are y = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow y=\pm \frac { 4 }{ \frac { 5 }{ 4 } } \Rightarrow y=\pm \frac { 16 }{ 5 } \)

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

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 = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow x=\frac { 5 }{ \frac { 13 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 13 } \)

\(\frac { { x }^{ 2 } }{ 3 } +\frac { { y }^{ 2 } }{ 10 } =1\)
Given equation \(\frac { { x }^{ 2 } }{ 3 } +\frac { { y }^{ 2 } }{ 10 } =1\)
This is an equation of the ellipse with major axis parallel to the y-axis.
∴ b² = 3, a² = 10
⇒ c² = a² - b² = 10 - 3
⇒ c = \(\sqrt { 7 } \)
\( a e=\sqrt{7} \)
\(\sqrt{10} =\sqrt{7} \)
\(e =\sqrt{\frac{7}{10}}\)
(a) Center is (0, 0)
⇒ h = 0, k = 0
(b) vertic are (h, k - a), (h, k + a)
⇒ (0, 0 - \(\sqrt { 10 } \)), (0, 0 + \(\sqrt { 10 } \))
⇒ (0 - \(\sqrt { 10 } \)), (0 + \(\sqrt { 10 } \))
(c) Foci are (h, k-c)(h, k+c)
⇒ (0, 0 - \(\sqrt {7 } \)), (0, 0 + \(\sqrt {7 } \))
Directrices are y = \(\pm \frac { a }{ e } \)
\(y=\pm \frac { \sqrt { 10 } }{ \frac { \sqrt { 7 } }{ \sqrt { 10 } } } \)
\(y=\pm \frac{10}{\sqrt{7}}\)

 y² - 4y - 8x + 12 = 0
y²-4y = 8x-12
Adding 4 both sides, we get,
y - 4y + 4 = 8x - 12 + 4 = 8x - 8
⇒ (y - 2)² = 8(x - 1)
This is a right open parabola and latus
rectum is 4a = 8 ⇒ a = 2.
(a) Vertex is (1, 2) ⇒ h = 1, k = 2
(b) focus is (h + a, 0 + k)
⇒ (1 + 2, 0 + 2)
⇒ (3, 2)
(c) Equation of directrix is x = h - a
⇒ x = 1-2
⇒ x = -1
(d) Length of latus rectum is 4a = 8 units.

 x²-2x+ 8y+ 17 = 0
x² - 2x = -8y - 17
Adding 1 both sides, we get
x² - 2x + 1 = -8y - 17 + 1
⇒ (x - 1)² = -8y - 16 = -8(y + 2)
⇒ (x - 1)² = -8(y + 2)
This is a open downward parabola, latus
rectum 4a = 8 ⇒ a = 2.
(a) Vertex is (1, -2)
⇒ h = 1, k = -2
(b) focus is (0 + h, - a + k)
⇒(0 + 1, -2-2)
⇒ (1, -4)
(c) Equation of directrix is y = k + a
⇒ y = -2 + 2 ⇒ y = 0
(d) Length of latus rectum is 4a = 8 units.

Let P(x, y) be the location of explosion
Given PB - PA = 6

Using distance formula,
\(\sqrt { { (x-5) }^{ 2 }+({ y-0) }^{ 2 } } -\sqrt { ({ x+5 })^{ 2 }+({ y-0) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { (x-5) }^{ 2 }+({ y })^{ 2 } } -\sqrt { (x+{ 5) }^{ 2 }+({ y) }^{ 2 } } =6\)
Squaring both sides, we get,
(x - 5)² + y² + (x + 5)² + y²
\(-2\sqrt { [(x-{ 5 })^{ 2 }+{ y }^{ 2 }][(x+{ 5 })^{ 2 }+{ y }^{ 2 }] } \) = 36

\(2\sqrt { [(x-{ 5 })^{ 2 }+{ y }^{ 2 }][(x+{ 5 })^{ 2 }+{ y }^{ 2 }] } \)
⇒ 2x² + 2y²+14
= \(2\sqrt { ({ x }^{ 2 }-10x+25+{ y }^{ 2 })({ x }^{ 2 }+10x+25+{ y }^{ 2 }) } \)
⇒ x² + y² + 7
= \(2\sqrt { ({ x }^{ 2 }-10x+25+{ y }^{ 2 })({ x }^{ 2 }+10x+25+{ y }^{ 2 }) } \)
Squaring again,
x⁴+ y⁴ + 49 +2x²y² + 14y² + 14x²
= (x² - 10x + 25 + y²) (x² + 10x + 25 + y²)

⇒ 14y² + 14x² = -50x² + 50y² + 625 - 49
⇒ 64x² - 36y² = 576
\(\div \) 4 we get,
16x² - 9y = 144
\(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\)
Hence the location of explo ion is restricted to a hyperbola whose equation is \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\)

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 \(\left( \frac { 4 }{ 3 } \right) \)

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 = \(\frac { 2\sqrt { 2 } }{ 3 } \) 

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.

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 the cross section of the satellite dish be an right open parabola.

Its equation is y² = 4ax
Since focus is placed 1.2 m from the vertex OA = 1.2 m and BC = 2.5 m since the width of the dish is 5 m.
From the diagram, a = 1.2 m
∴ y² = 4(1.2)x
(a) ⇒ y² = 4.8x   ...(1)
(b) Since (x₁, 2.5) lines on (1)(2.5)² = 4.8(x₁)
x₁ = \(\frac { 2.5\times 2.5 }{ 4.8 } \)
x₁ = 1.3 m
∴ Depth of the satellite dish at the vertex is 1.3 m.

Let the equation of the parabola be
(x - h)² = -4a(y - k).
Here the vertex is (0.5, 4)
Equation of the parabola is (x - 0.5)²
= -4a(y-4) ...(1)
O(0, 0) is a point on the parabola
(0 - 0.5)² = -4a (0 - 4)
⇒ \({ \left( \frac { -1 }{ 2 } \right) }^{ 2 }=-4a(-4)\)
⇒ \(\frac { 1 }{ 4 } =16a\Rightarrow a=\frac { 1 }{ 64 } \)
∴ (1) becomes as (x - 0.5)² = \(-4\times \frac { 1 }{ 64 } (y-4)\)
Also D(0.75, y₁) is a point on the parabola
∴ (0.75 - 0.5)² = \(\frac { -1 }{ 16 } ({ y }_{ 1 }-4)\)
⇒ \({ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -1 }{ 6 } ({ y }_{ 1 }-4)\)
\(\Rightarrow \frac{1}{\not 16}=\frac{1}{\not16}\left(y_{1}-4\right)\)
⇒ 1 = -y₁ + 4
⇒  y₁ = -1 + 4 = 3m
Height of the water at a horizontal distance of 0.75m is 3m

Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Length of semi minor axis b = 5
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Let BB' be the road width and AA' be the end points of the opening of the tunnel.
Let CB = 8, BD = 1,
D is (8, 4) lies on the ellipse
\( \frac{8^{2}}{a^{2}}+\frac{4^{2}}{5^{2}}=1 \)
\( \Rightarrow a^{2}=\frac{25}{9} \times 64 \)
\( \Rightarrow a=\frac{40}{3} \)
The width AA' = 2a
\(=\frac{80}{3}=26.66 \mathrm{~m}\)
The required width is 26.66 m.

Since the centre is located at (0, 300), midway between the two foci, which are the coast guard stations, the equation is \(\frac { { \left( y-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { { \left( x-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\).... (1)
To determine the values of a and b, select two points known to be on the hyperbola and substitute each point in the above equation.
The point(0, 400) lies on the hyperbola, since it is 200 km further from Station A than from station B.
\(\frac { { \left( 400-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { O }{ { b }^{ 2 } } =1\frac { { 100 }^{ 2 } }{ a^{ 2 } } =1,{ a }^{ 2 }=10000.\) There is also a point (x, 600) on the hyperbola such that 600²+ x² = (x + 200)²    
360000 + x² = x²+400x+40000
x = 800
Substituting in (1), we have \(\frac { { \left( 600-300 \right) }^{ 2 } }{ 10000 } -\frac { { \left( 800-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\)
\(9-\frac { 640000 }{ { b }^{ 2 } } =1\)
b² = 80000
Thus the required equation of the hyperbola is \(\frac { { \left( y-300 \right) }^{ 2 } }{ 10000 } -\frac { { x }^{ 2 } }{ 80000 } =1\)
The ship lies somewhere on this hyperbola. The exact location can be determined using data from a third station.

Since the truck’s width is 3m, to determine the clearance, we must find the height of the archway 1.5 m from the centre. If this height is 2.7 m or less the truck will not clear the archway.

From the diagram a = 6 and b = 3 yielding the equation of ellipse as \(\frac { { x }^{ 2 } }{ { 6 }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 }^{ 2 } } =1\)
The edge of the 3m wide truck corresponds to x = 1.5 m. We will find the height of the
archway 1.5 m from the centre by substituting x =1.5 and solving for y 
\(\frac { { \left( \frac { 3 }{ 2 } \right) }^{ 2 } }{ 36 } \frac { { y }^{ 2 } }{ 9 } =1\)
 \({ y }^{ 2 }=9\left( 1-\frac { 9 }{ 144 } \right) \)
\(\frac { 9\left( 135 \right) }{ 144 } =\frac { 135 }{ 16 } \)
\(y=\frac { \sqrt { 135 } }{ 4 } \)
= 2.90
Thus the height of arch way 1.5m from the centre is approximately 2.90m. Since the truck’s height is 2.7 m, the truck will clear the archway.

Equation of normal at 't₁' to the parabola y² = 4 axis is
y + xt₁ = at₁³ + 2at₁ ...(1)
(1) meets the-parabola y₂ = 4ax at 't₂'.
At 't₂', the point on the parabola is x = at²₂, y = 2at₂ (2) lies on (1)
∴ Substituting (2) in (1) we get,
2at₂ + (at₂²)2t₁ = at₁³ + 2at₁
⇒ 2at₂ + at₁t₂² = at₁³ + 2at₁
⇒ 2a(t₂ - t₁) = -at₁[t₂² - t₁²]

⇒ 2 = -t₁(t₂+ t₁)
⇒ \(\frac { -2 }{ { t }_{ 1 } } ={ t }_{ 2 }+{ t }_{ 1 }\Rightarrow { t }_{ 2 }=\frac { -2 }{ { t }_{ 1 } } { -t }_{ 1 }\)
⇒ t₂ = -(t₁ + \(\frac { -2 }{ { t }_{ 1 } } \))
Hence proved.

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)]

Equation of the parabola is y² = 8x
∴ 4a = 8 ⇒ a = 2
Equation of tangent to the parabola in parametric form is yt = x + at²
When t = 2, the equation of tangent is
y(2) = x + 2(2)² ⇒ 2y = x + 8
⇒ x - 2y + 8 = 0 is the required equation of tangent.

Equation of theparabola is y² = 16x
∴ 4a = 16 ⇒ a = 4
Let y = mx + c ... (1)
be a tangent to the parabola
Since the tangent is perpendicular to
2x + 2y+ 3 =0,
m = \(\frac{-1}{Slope \ of \ the \ line \ 2 x + 2y + 3 = 0}\)
∴ m = \(\frac { -1 }{ \frac { -2 }{ 2 } } =\frac { -1 }{ -1 } \) = 1
[∵ \(Slope=\frac{Co - efficient of \ x}{Co - efficient of \ y}=\frac{-2}{2}\) = -1]
The condition for the line y = mx + c to be a tangent to the parabola is c = \(\frac{a}{m}\)
∴ c = \(\frac14\)= 4
From (1), y = 1(x) + 4 ⇒ x - y + 4 = 0 is the required tangent.

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).

Given equation of hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 64 } =1\)
Let y = mx + c be the required tangent
⇒ a² = 16 and b² = 64
The tangents are parallel to 10x - 3y + 9 = 0.
∴ Slope of tangent (m)
= Slope of the line 10x - 3y + 9 = 0.
∴ m = \(\frac{-co - efficient of \ x}{co - efficient of \ y}\)= \(\frac { -10 }{ -3 } =\frac { 10 }{ 3 } \)
The condition for y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
⇒ \({ c }^{ 2 }=16\left( \frac { 100 }{ 9 } \right) -64=\frac { 1600-64\times 9 }{ 9 } \)
⇒ \(\frac { 1024 }{ 9 } \Rightarrow c=\pm \frac { 32 }{ 3 } \)
∴ The required tangents are
y = \(\frac { 10 }{ 3 } x+\frac { 32 }{ 3 } \) or \(y=\frac { 10x }{ 3 } -\frac { 32 }{ 3 } \)
⇒ 3y = 10x + 32 or 3y = 10x - 32
⇒ 10x−3y+32 = 0, or 10x+3y−32 = 0

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

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 

\(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
Given equation is \(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
This is an equation of the ellipse a² = 289
b² = 225 and
c² = a² - b² ⇒ 289 - 225 = 64 ⇒ c = 8.
\(e=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 225 }{ 289 } } =\sqrt { \frac { 289-225 }{ 289 } } \)
 \(=\sqrt { \frac { 64 }{ 289 } } =\frac { 8 }{ 17 } \)
(a) Center is (3, 4) ⇒ h = 3, k = 4
(b) foci are (h, k+c), (h, k-c)
⇒ (3, 4 + 8), (3, 4 - 8) ⇒ (3, 12), (3,-4)
(c) Vertices are (h, k - a), (h, k + a)
⇒ (3, 4 -17), (3, 4 + 17) ⇒ (3, -13), (3, 21)
(d) Equations of directrices are y - 4 = \(\pm \frac { a }{ e } \)
\(\Rightarrow y-4=\pm \frac { 17 }{ \frac { 8 }{ 17 } } +4\Rightarrow y-4=\pm \frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289 }{ 8 } +4\) and \(y=\frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289+32 }{ 2 } \) and \(y=\frac { -289+32 }{ 8 } \)
\(\Rightarrow y=\frac { 321 }{ 8 } \) and \(y=\frac { -257 }{ 8 } \)

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
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 = \(\pm \frac { 5 }{ \frac { 4 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 4 } \)

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.

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.

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.

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\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ........... (1)
(1) passes through (1, 0)
⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0
⇒ 2g + c = -1 ................(2)
(1) passes through (-1, 0)
⇒ (-1)² + 0 + 2g(-1) + 2f(0)+ c = 0
⇒ -2g + c = -1 ..............(3)
Also (1) passes through (0, 1)
⇒ 0 + 1²+ 2g(0) +2f(1) + c = 0
⇒ 2f + c = -1 .................(4)
(2) + (3) ⇒ 2c = -2
⇒ c = -1
Substituting c = -1 in (2), we get
2g-1 = -1
⇒ 2g = 0
⇒ g = 0
Substituting c = -1 in (4) we get,
2f -1 = -1
⇒ 2f  = 0
⇒ f = 0
∴ The required equation of the circle 
x² + y² - 1 = 0

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.

The length a of the semi major axis of the elliptical ceiling is17 m. The height b of the semi minor axis is 8 m. Thus c² = a² -b² = 17² - 8²
then c =\(\sqrt { 289-64 } =\sqrt { 225 } =15\)
For the elliptical ceiling the foci are located on either side about 15 m from the centre, along its major axis.

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.

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Let P(x, y) be any point on the hyperbola
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)

Then by definition, SP = ePM and S'P = ePM'
SP = ePM ⇒ SP = e(NK)
= e(CN - CK)
= \(e\left( x-\frac { a }{ e } \right) \)  = ex - a
= and S'P = ePM' ⇒ S'P = e(NK')
⇒ e(CN + CK') = \(e\left( x+\frac { a }{ e } \right) \) = ex + a
∴ S'P -SP = (ex + a) - (ex - a)
ex + a - ex + a = 2a (constant)
= length of the transverse axis.

The latus rectum LL' of the hyperbola passes through S(ae, 0)

∴ L is (ae, y₁)
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \) = 1 we get,
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } -\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow { e }^{ 2 }-\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow { e }^{ 2 }-1=\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } \) \([{ b }^{ 2 }={ a }^{ 2 }({ e }^{ 2 }-1)\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }({ e }^{ 2 }-1)\) \(\Rightarrow \frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1]\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \) \(\Rightarrow { { y }_{ 1 } }^{ 2 }=\frac { { b }^{ 4 } }{ { a }^{ 2 } } \)
\(\Rightarrow { y }_{ 1 }=\pm \frac { { b }^{ 2 } }{ a } \)
∴ End points oflatus rectum Land L' are
\(\left( ae,\frac { { b }^{ 2 } }{ a } \right) \) and \(\left( ae,-\frac { { b }^{ 2 } }{ a } \right) \)
Hence, the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } =\frac { 2{ b }^{ 2 } }{ a } \) units.
Hence proved.

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.

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 

Given equation of the circle is
x² + y² - 5x + 2y - 5 = 0
(i) At (-2, 1), (1) becomes
(-2)² + 1²- 5(-2) + 2(1) - 5
= 4 + 1 + 10 + 2 - 5
= 17 - 5 = 12 > 0
∴ (-2, 1) lies outside the circle.
(ii) At (0, 0), (1) becomes -5 < 0
∴ (0, 0) lies inside the circle.
(iii) At (-4, -3), (1) becomes
(-4)² + (-3)² - 5(-4) + 2(-3) - 5
= 16 + 9 + 20 - 6 - 5
= 45 - 11 = 34 > 0
∴ (-4, -3) lies outside the circle.