Let P(x, y) be a moving point. By definition
\( \frac{F P}{P M}=\mathrm{e} \)
\( \therefore F P^{2}=e^{2} P M^{2} \)
\( (x+1)^{2}+(y+3)^{2}=\frac{16}{25}\left[\pm \frac{x-2 y}{\sqrt{1+4}}\right]^{2} \)
\( 125\left[(x+1)^{2}+(y+3)^{2}\right]=16(x-2 y)^{2} \)
\(\Rightarrow 109 x^{2}+64 x y+61 y^{2}+250 x+750 y+1250=0\)

Let P(x, y) be a moving point. By definition
\( \frac{F P}{P M} =e \)
\(\therefore F P^{2} =e^{2} P M^{2} \)
\( (x-2)^{2}+(y-0)^{2} =\frac{1}{4}\left(\pm \frac{x-8}{\sqrt{1}}\right)^{2} \)
\((x-2)^{2}+y^{2} =\frac{1}{4}(x-8)^{2} \)
\( 4\left[(x-2)^{2}+y^{2}\right] =(x-8)^{2} \)
\(3 x^{2}+4 y^{2} =48 \)
\(\frac{x^{2}}{16}+\frac{y^{2}}{12} =1 \)
Aliter:

From the given data, the major axis is along the x-axis and the equation of the ellipse may be taken as
\( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \)
\(F Z=\frac{a}{e}-a e=6 \)
\( e=\frac{1}{2} \Rightarrow 2 a-\frac{1}{2} a=6 \)
\(\Rightarrow \frac{3}{2} a=6 \Rightarrow a=4 \)
\( b^{2}=a^{2}\left(1-e^{2}\right)=16\left(1-\frac{1}{4}\right)\)
\( =16 \times \frac{3}{4}=12 \)
Therefore, The required equation is \(\frac{x^{2}}{16}+\frac{y^{2}}{12} =1\)

\(\frac{(x+3)^{2}}{6}+\frac{(y-5)^{2}}{4}=1\)
The major axis is parallel to x axis.
Centre (h, k) = (-3, 5)
\( a^{2}=6 \quad b^{2}=4 \)
\(a=\sqrt{6} \quad b=2 \)
\( c^{2}=a^{2}-b^{2}=6-4=2 \)
\( c=\sqrt{2} \)
\( a e=\sqrt{2} \)
\( \sqrt{6} e=\sqrt{2} \)
\( e=\frac{1}{\sqrt{3}}\)
Vertices \( (h \pm a, k) =(-3 \pm \sqrt{6}, 5) \)
\( =(-3+\sqrt{6}, 5) \&(-3-\sqrt{6}, 5)\)
foci \( (h \pm c, k) =(-3 \pm \sqrt{2}, 5) \)
\( =(-3+\sqrt{2}, 5) \&(-3-\sqrt{2}, 5) \)

According to the problem, the ordinates of vertex and focus are equal hence, the axis of the required parabola is parallel to x- axis.
Again,
a = abscissa of focus - abscissa of vertex
⇒ a = 1-(- 2) = 1+2 = 3
Therefore, the equation of the required parabola is
\( (y-k)^{2} =4 a(x-h) \)
\(\Rightarrow(y-3)^{2} =4.3(x+2) \)
\(\Rightarrow y^{2}-6 y+9 =12 x+24 \)
\(\Rightarrow y^{2}-6 y-12 x-15 =0 \)

The given parabola is y² = 12x.
Now, let (k, 2k) be the co-ordinates of the required point (k \(\neq\)0).
Since the point lies (k, 2k) on the parabola y² = 12x,
Therefore we get
\( (2 k)^{2} =12 k \)
\(\Rightarrow 4 k^{2} =12 k \)
\(\Rightarrow k =3(\text { since, } k \neq 0) \)
Therefore, the co-ordinates of the required point are (3, 6).

The equation of the parabola is
\(x^{2}+x-2 y+2=0\)
Equation of the tangent at (x₁, y₁) to the given parabola is
\( x x_{1}+\frac{x+x_{1}}{2}-2 \frac{\left(y+y_{1}\right)}{2}+2=0 \)
\( \text {At }(1,2), \ \ x(1)+\frac{x+1}{2}-2 \frac{(y+2)}{2}+2=0 \)
On simplification we get 3x - 2y + 1 = 0
Equation of the normal is of the form 2x+3y+k = 0
This normal passes through (1, 2)
\(\therefore\) 2 + 6 + k = 0
⇒ k = -8
\(\therefore\) Equation of the normal is 2x + 3y - 8 = 0

a. centre at the orign and radius 7.
Equation of the circle
\( x^{2}+y^{2}=r \)
\( x^{2}+y^{2}=7^{2} \)
\( x^{2}+y^{2}=49\)
b. Centre at (-1, 3) and radius 4.
Centre (h, k) = (-1, 3)
Radius r = 4
Equation of the circle
\( (x-h)^{2}+(y-k)^{2}=r^{2} \)
\((x+1)^{2}+(\mathrm{y}-3)^{2}=4^{2} \)
\(x^{2}+2 x+1+y^{2}-6 y+9-16=0 \)
\(x^{2}+y^{2}+2 x-6 y-6=0 \)
c. Centre at (3, 1) and passing through the point (5, 4).
Centre C (h, k) = (3,1)

r = CP
\( =\sqrt{(5-3)^{2}+(4-1)^{2}} \)
\(=\sqrt{2^{2}+3^{2}} \)
\(=\sqrt{4+9}=\sqrt{13} \)
Equation of circle
\( (x-h)^{2}+(y-k)^{2} =r^{2} \)
\((x-3)^{2}+(y-1)^{2} =(\sqrt{13})^{2} \)
\(x^{2}-6 x+9+y^{2}-2 y+1 =13 \)
\(x^{2}+y^{2}-6 x-2 y-3 =0 \)

The condition for the tangency is
\(c^{2}=a^{2}\left(1+m^{2}\right)\)
Here \(a^{2}=16, m=-\frac{3}{4}, c=\frac{p}{4}\)
\( c^{2}=a^{2}\left(1+m^{2}\right) \Rightarrow \frac{p^{2}}{16} =16\left(1+\frac{9}{16}\right) \)
\(\frac{p^{2}}{\not16} =16 \times \frac{25}{\not16} \)
\(p^{2} =16 \times 25 \)
\(\therefore p =\pm 20\)

To find the end points of the chord, solve the equations of the circle and the line.
Substitute x = 2y + 1 in the equation of the circle
\( (2 y+1)^{2}+y^{2}-2(2 y+1)-y+1 =0 \)
\(4 y^{2}+4 y+1+y^{2}-4 y-2-y+1 =0 \)
\(5 y^{2}-y =0 \)
\(\therefore y(5 y-1) =0 \ \ \)
y = 0 \(\text { (or) } y =\frac{1}{5} \)
\(\Rightarrow x =1 \text { (or) } x =\frac{7}{5}\)
The two end points are (1, 0) and \(\frac{7}{5}, \frac{1}{5}\)
Length of the chord \(=\sqrt{\left(1-\frac{7}{5}\right)+\left(0-\frac{1}{5}\right)^{2}}\)
\(=\sqrt{\frac{4}{25}+\frac{1}{25}}=\frac{1}{\sqrt{5}}\) units

The equation of the tangent at (x₁, y₁) to the given circle is
\( x x_{1}+y y_{1}-4\left(\frac{x+x_{1}}{2}\right)+4\left(\frac{y+y_{1}}{2}\right)-8=0 \)
\( x x_{1}+y y_{1}-2\left(x+x_{i}\right)+2\left(y+y_{1}\right)-8=0 \)
At (1, -2), the equation of the tangent is
-2x - 2y - 2(x - 2) + 2(y - 2) - 8 = 0
\(\Rightarrow\) -4x - 8 = 0
\(\Rightarrow\) x + 2 = 0 is the required equation of the tangent.

 The given circle is \(x^{2}+y^{2}-4 x+6 y-9=0\)
Centre (- g, -f) is (2, 3)
The circle passes through the point (- 4, - 5).
\(\therefore\) radius \( =\sqrt{(2+4)^{2}+(3+5)^{2}}=\sqrt{36+64} \)
\(=\sqrt{100}=10 \)
The equation of the circle is
\( (x-h)^{2}+(y-k)^{2}=r^{2} \)
Here \( (h, k)=(2,3), r=10 \)
\( \therefore(x-2)^{2}+(y-3)^{2}=10^{2}\)
\(x^{2}+y^{2}-4 x-6 y-9=0\) is the required equation of the circle.

Given line is x + y + 1 = 0
⇒ y = -x-1
m = -1, c = -1
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
a² = 16, b² = 15
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ (-1)² = 16(-1)² - 15
1 = 16 -15
1 = 1
Since the condition is satisfied, x + y + 1 = 0 touches the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 15 } \) = 1
The point of contact is \(\left( \frac { -{ a }^{ 2 }m }{ c } ,\frac { -{ b }^{ 2 } }{ c } \right) \) = \(\left( \frac { -16(-1) }{ -1 } ,\frac { -15 }{ -1 } \right) \) = (-16, 15)
Hence, the point of contact is (-16, 15)

Given line is y = x + c
m = 1, c = c
Equation of the hyperbola is 9x² - 16y² = 144
\(\div \)144 we get \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
a² = 16, b² = 9
The condition for the line y = mx + c to be a tangent to the hyperbola is c² = a²m² - b²
∴ c² = 16(1)² - 9 = 16 - 9 = 7
∴ c = ±\(\sqrt7\)

Given equation of the hyperbola is
3x² - 6y² = -18; \(\div \)by (-18) we get \(\frac{y^2}{3}- \frac{x^2}{6}\) = 1
The transverse axis is long y-axis.
Here a² = 3, b² = 6
Length of transverse axis is 2a = 2\(\sqrt { 3 } \)
Length of conjugate axis is 2b = 2\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

Given 2b = 5 and 2ae = 13
b² = a²( e² - 1) - b ⇒ a \(\sqrt { { e }^{ 2 }-1 } \)
2b = 5 ⇒ 2a\(\sqrt { { e }^{ 2 }-1 } \) = 5
⇒ 4a²( e² - 1) = 25 [squaring both sides]
⇒ 4a²e²- 4a² = 25
⇒ (2ae)² - 4a² = 25
⇒ 13²-4a²=25 [∵ 2ae=13]
⇒169- 25 = 4a²
⇒ 4a²= 144
⇒ a²= 36
⇒ a = 6
∴ 2b = 5 ⇒ b = \(\frac52\)⇒b² = \(\frac{25}{4}\)
∴ Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ 36 } -\frac { { y }^{ 2 } }{ \frac { 25 }{ 4 } } =1\)
\(\frac { { x }^{ 2 } }{ 36 } -\frac { { 4y }^{ 2 } }{ 25 } =1\)