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 = \(\frac { 5 }{ 4 } \)

Let the equation of the ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given, length of LR = \(\frac { 1 }{ 2 } \) (Length of major axis)
⇒ \(\frac { { 2b }^{ 2 } }{ a } =\frac { 1 }{ 2 } \) (2a) ⇒ \(\frac { 2{ b }^{ 2 } }{ a } \) = a ⇒ 2b² = a²
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { { b }^{ 2 } }{ { 2a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { 1 }{ \sqrt { 2 } } \)

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 = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

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.

Let AOB be the vertical section of the reflector and m is the mid-point of AB. Let the equation of the parabola be y² = 4ax A(6, 12) lies on (1)

∴ 12² = 4a(6) ⇒ a = 6
∴ Focus is (a, 0) = (b, 0)
Hence focus coincides with m, the mid-point of AB.

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 - 8x - 9y + 12 = 0
Length of the tangent = \(\sqrt { { 2 }^{ 2 }+({ -3) }^{ 2 }-8(2)-9(-3)+12 } \)
= \(\sqrt { 4+9-16+27+12 } \)
= \(\sqrt { 36 } \) = 6 unit

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 + 6x - 4y + 4 = 0.
Here 2g = 6 ⇒ g = 3
2f  = -4 ⇒ f = -2 and c = 4
Centre is (-g, -f) ⇒ (-3, 2)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { 3 }^{ 2 }+(-2){ }^{ 2 }-{ 4 } } \)
= \(\sqrt { 9+4-4 } \)
= \(\sqrt { 9 } \)
= 3 unit

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.

Taking (x₁, y₁) as (2, 3), we get
x₁² + y₁² + 2gx₁+ 2fy₁+ c = 2² + 3² − 6 × 2 − 8 × 3 + 12 
= 4 + 9 - 12 - 24 + 12
= -11\(<\)0.
Therefore the point(2, 3) lies inside the circle, by theorem.

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