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 \( y=-2-3\sqrt { 6 } \)

 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 \(\left( \frac { 4 }{ 3 } \right) \)