Let the parabola be y² = 4ax 
Since focus is 2m from the vertex a = 2
Equation of the parabola is y² = 8x
Let P be a point on the parabola whose x -coordinate is 3m from the
vertex P (3, y) 
y² = 8 × 3
y =\(\sqrt { 8\times 3 } \)
= \(2\sqrt { 6 } \)
The width of the antenna 3m from the vertex is 4\(\sqrt { 6 } \) m.

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

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.

A = 11, C = -25, D = -44, E = - 50, and F = -256
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

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 \( y=-2-3\sqrt { 6 } \)

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