The major axis is along x-axis
Here \(a^{2}=16,\ b^{2}=9\)
\(e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{10}}=\frac{\sqrt{7}}{4}\)
Equations of directrices are x = \(\pm \frac{a}{e}\)
\(x=\frac{\pm 16}{\sqrt{7}}\)
Equations of the latus rectums are x = \(\pm\)ae
\(x=\pm \sqrt{7} \)
Length of the latus rectum \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\)

Let x - 1 = X and y + 1 = Y
\(\therefore\) The given equation becomes \(\frac{X^{2}}{9}+\frac{Y^{2}}{16}=1\)
Clearly the major axis is along Y-axis and the minor axis is along X-axis. 
The equation of major axis is X = 0 and the equation of minor axis is Y = 0.
i.e., the equation of major axis is x - 1 = 0 and the equation of minor axis is y + 1 = 0
Here \(a^{2}=16,\ b^{2}=9\)
\(\therefore\) Length of major axis (2a) = 3  
\(\therefore\) Length of minor axis (2b) = 6

\(\frac{x^{2}}{3}+\frac{y^{2}}{4}=1\)
The major axis is along y-axis and the minor axis is along x-axis.
\(\therefore\) The equation of major axis is x = 0 and the equation of minor axis is y = 0 Here a² = 4;
\(b^{2}=3 \Rightarrow a=2, b=\sqrt{3}\)
The length of major axis (2a) = 4
The length of minor axis (2b) = 2\(\sqrt3\)

The major is along x-axis and the minor axis is along y-axis.
This gives the equation of major axis as y = 0 and equation of minor axis as x = 0.
We have a² = 9, b² = 4 ⇒ a = 3, b = 2.
\(\therefore\) The length of major axis is 2a = 6 and the length of minor axis is 2b = 4

The given equation of the parabola is \((x+2)^{2}=-4(y+1)\).
Then parametric equation of the parabola (x+2)² = -4(y + 1)are
x + 2 = 2t and y+1 = -t²
x = 2t - 2 and y = -t²-1.

The equation of chord of contact of tangents from (x₁, y₁) to 2x² + 5y² - 20 = 0 is \(2 x x_{1}+5 y y_{1}-20=0\)
\(\therefore\) the required equation from (2, 4) is
2x(2)+5y(4)-20 = 0
i.e. x+5y-5 = 0

The condition for the line y = mx+c to be a tangent to
\(x^{2}+y^{2}=a^{2} \text { is } \mathrm{c}=\pm a \sqrt{1+m^{2}}\)
Here a = 3, m = 3
\(\therefore c=\pm 3 \sqrt{10}\)

The equation of the circle is \(x^{2}+y^{2}=25\)
The equation of the tangent at (x₁, y₁) is \(x x_{1}+y y_{1}=25 .\)
Here (x₁, y₁) = (4, 3).
\(\therefore\) The equation of the tangent at (4, 3) is 4x+3y = 25

The length of the tangent PT from P(x₁, y₁) to the circle
\( x^{2}+y^{2}+2 g x+2 f y+c=0 \) is 
\(\mathrm{PT}=\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c} \)
\( \mathrm{PT}^{2} =2^{2}+3^{2}-6.2-8.3+12 \)
\( =4+9-12-24+12 \)
\( =-11<0 \)
The point (2, 3) lies inside the circle

The length of the tangent to the circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) from the point 
\(\left(x_{1}, y_{1}\right) \text { is } \sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent to the given circle
\( =\sqrt{x_{1}^{2}+y_{1}^{2}-4 x_{1}-3 y_{1}+12} \)
\(=\sqrt{2^{2}+3^{2}-4.2-3.3+12} \)
\(=\sqrt{4+9-8-9+12} \)
\(=\sqrt{8} \)
\(=2 \sqrt{2} \text { units } \)

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 \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)