Let equation of the hyperbola be
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
\(Now\quad \frac { { 2b }^{ 2 } }{ a } =8\quad \Rightarrow \quad { b }^{ 2 }=4a\)
\(But\quad { b }^{ 2 }={ a }^{ 2 }({ e }^{ 2 }-1)\)
\(\therefore { a }^{ 2 }({ e }^{ 2 }-1)=4a\quad \Rightarrow \quad a({ e }^{ 2 }-1)=4\)
\(\Rightarrow a\left( \frac { a }{ 5 } -1 \right) =4\quad \Rightarrow \quad a=5\)
\(\therefore \quad { b }^{ 2 }=4\times 5=20\)
\(Thus\quad equation\quad of\quad required\quad hyperbola\quad is\)
\(\frac { { x }^{ 2 } }{ 25 } -\frac { { y }^{ 2 } }{ 20 } =1\)

\(Let\quad equation\quad of\quad elipse\quad be\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(The\quad co-ordinates\quad of\quad foci\quad are(\pm ae,0).\)
\(\therefore ae=2\quad \Rightarrow \quad a\times \frac { 1 }{ 3 } =2\)
\(\Rightarrow \quad a=6\)
\( Now\quad { b }^{ 2 }={ a }^{ 2 }\left( { 1-e }^{ 2 } \right) \)
\(Thus\quad equation\quad of\quad required\quad elipse\quad is\)
\(\frac { x^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 32 } =1\)

\(\text{ Let equation of ellipse be }\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)

Since the vertices of the elipse are an y − axis
∴ b=10
\( Now\quad { a }^{ 2 }={ b }^{ 2 }(1-{ c }^{ 2 })\)
\(\Rightarrow { a }^{ 2 }={ \left( 10 \right) }^{ 2 }\left[ 1-\left( \frac { 3 }{ 3 } \right) ^{ 2 } \right] -100\times \frac { 16 }{ 25 } =64\)
Thus equation of required elipse is
\(\frac { { x }^{ 2 } }{ 64 } +\frac { y^{ 2 } }{ 100 } =1\)

Let ABC be an equilateral triangle inscribed in the circle
x²+y²+2gx+2fy+c=0.
Then center of circle isO(−g,−f).
\(\therefore OA=OB=OC=\sqrt { { { g }^{ 2 }{ +f }^{ 2 }-c } } \)
\(In\triangle OBD\)
\(\sin { 60^{ 0 } } =\frac { BD }{ OB } \Rightarrow \frac { \sqrt { 3 } }{ 2 } =\frac { BD }{ \sqrt { { g }^{ 2 }{ f }^{ 2 }-c } } \)
\(\therefore \quad BD=\frac { \sqrt { 3 } }{ 2 } \sqrt { { g }^{ 2 }{ f }^{ 2 }-c } \)

\(Also\quad BC=2BD\)
\( \Rightarrow BC=\sqrt { 3 } \sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\( Area\quad \triangle ABC=\frac { \sqrt { 3 } }{ 4 } \times \left( BC \right) ^{ 2 }\)
\(=\frac { \sqrt { 3 } }{ 4 } \times 3({ g }^{ 2 }+{ f }^{ 2 }-c)\)
\(=\frac { 3\sqrt { 3 } }{ 4 } { (g }^{ 2 }{ +f }^{ 2 }-c)\) sq.units

we know that the center of the circule touching the co−ordinate axis are
(a,a),where a is the radius of the circle.
The equation of circule is
(x−a)²+(y−a)²=a2
⇒x²+y²−2ax−2ay+a²=0
Also the center (a,a) lies on the line
lx+my+n=0
∴la+ma+n=0⇒ \( a=-\frac { n }{ 1+m } \)
\(Putting\quad this\quad value\quad of\quad a\quad in(i),\quad we\quad have\)
\({ x }^{ 2 }+y^{ 2 }+\frac { 2nx }{ l+m } +\frac { 2ny }{ l+m } +\frac { { n }^{ 2 } }{ (l+m)^{ 2 } } =0\)
\(\Rightarrow { \left( l+m \right) }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 })+2n(l+m)(x+y)+{ n }^{ 2 }=0\)

The equation of given circule is x2+y2−2x−4y−20=0
whose centere is(1,2) and radius is 5.
Now this circles touches another circle externally at point (3,7).
Let (a,b) be the center of required circle then(3,7) is the mid point of line segment joining the centres of two circles.
\(\therefore \frac { a+1 }{ 2 } =3\quad and\quad \frac { b+2 }{ 2 } =7\)
\(\therefore \quad a=5\quad and\quad b=12\)
thus equation of required circle is
\(\left( { x-5 } \right) ^{ 2 }+\left( y-12 \right) ^{ 2 }=\left( 5 \right) ^{ 2 }\)

Here width of elliptical arch = 8 m
∴ AB=8m ⇒ 2a=8 ⇒ a=4
Height at the centre = 2 m
∴ OB= 2 ⇒ b = 2
The axis of the ellipse is x-axis.
So the equation of ellipse in standard form is
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ \(\frac { { x }^{ 2 } }{ { (4) }^{ 2 } } +\frac { { y }^{ 2 } }{ (2)^{ 2 } } \)=1 ⇒ \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } \)=1
Now AP=1.5 m
∴ OP = OA - AP = 4 - 1.5 = 2.5 m
Let PQ = h
∴ Coordinates of Q are (2.5, h)
Since the point Q lies on the ellipse
\(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } \)=1
∴ \(\frac { (2.5)^{ 2 } }{ 16 } +\frac { { h }^{ 2 } }{ 4 } \)=1 ⇒ \(\frac { { h }^{ 2 } }{ 4 } =\frac { 1-6.25 }{ 16 } \)
⇒ h²=\(\frac { 9.75\times 4 }{ 16 } =\frac { 9.75 }{ 4 } \)
⇒ h²=2.44
⇒ h=\(\sqrt2.44\).

Here foci are (± 3\(\sqrt5\), 0) which lie on x-axis.
So the equation of hyperbola in standard form is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ foci (±c, 0) is (± 3\(\sqrt5\), 0) ⇒ c = 3\(\sqrt5\)
Length of latus rectum \(\frac { 2{ b }^{ 2 } }{ a } \)=8 ⇒ b2=4a
We know that c² = a² + b²
∴ (3\(\sqrt5\))² = a² + 4a  ⇒ a² + 4a - 45 = 0
⇒ (a + 9) (a - 5) = 0 ⇒ a = -9 or a = 5
(∵ a = -9 is not possible)
Also ⇒ a = 5 ⇒ a² = 25b² = 4 x 5 = 20
Thus required equation of hyperbola is
\(\frac { { x }^{ 2 } }{ 25 } -\frac { { y }^{ 2 } }{ 20 } \)=1

The vertices are (0, ± 5) which lie on y-axis.
So the equation of the hyperbola in standard form is \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ the vertices (0, ± a) is (0, ± 5) ⇒ a = 5
foci (0, ± ae) is (0, ± 8) ⇒ ae = 8
Now ae=8 ⇒ e=\(\frac { 8 }{ a } \) ⇒ \(\frac { 8 }{ 5 } \)
We know that b =a\(\sqrt { { e }^{ 2 }-1 } \)
⇒ e=\(5\sqrt { \frac { 64 }{ 25 } -1 } =5\frac { \sqrt { 39 } }{ 5 } =\sqrt { 39 } \)
Thus required equation of hyperbola is
\(\frac { { y }^{ 2 } }{ (5)^{ 2 } } -\frac { { x }^{ 2 } }{ (\sqrt { 39 } )^{ 2 } } \)=1 ⇒ \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 39 } \)=1

The vertices are (±2, 0) which lie on x-axis
So the equation of hyperbola in standard form is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ the vertices (± a, 0) is (± 2, 0) ⇒ a =2
foci (± ae, 0) is (± 3, 0) ⇒  ae = 3
Now ae=3 ⇒ e=\(\frac { 3 }{ a } \) ⇒ e=\(\frac { 3 }{ 2 } \)
we know that b=a\(\sqrt { { e }^{ 2 }-1 } \)
⇒ b=2\(\sqrt { \frac { 9 }{ 4 } -1 } =2\frac { \sqrt { 5 } }{ 2 } =\sqrt { 5 } \)
Thus required equation of hyperbola is
\(\frac { { x }^{ 2 } }{ (2)^{ 2 } } -\frac { { y }^{ 2 } }{ (\sqrt { 5 } )^{ 2 } } \) =1 ⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 5 } \)=1

Directrix is x = -2 and the focus is (2, 0).
Required equation of parabola is y² = 4ax
y² = 4 X 2x    [a =2]
y² = 8x

We nave, the major axis of the ellipse lies on the y-axis . so the equation of an ellipse is
\(\frac { { x }^{ 2 } }{ { b }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1,a>b\)
Since,the ellipse passes through the point (3,2) and (1,6).
\(\therefore \frac { 9 }{ { b }^{ 2 } } +\frac { 4 }{ { b }^{ 2 } } =1\)
\(and\quad \frac { 1 }{ { b }^{ 2 } } +\frac { 36 }{ { a }^{ 2 } } =1\)
On solving eqs.(i) and (ii),we get a²=40,b²=10
hence,the equation of ellipse is \( \frac { { x }^{ 2 } }{ 10 } +\frac { { y }^{ 2 } }{ 40 } =1.\)

We have, conjugate axis is half of distance foci.
\(2b=\frac { 1 }{ 2 } 2c\quad \Rightarrow \quad 2b=c\quad \Rightarrow \quad { 4b }^{ 2 }={ c }^{ 2 }\)
\({ 4b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\Rightarrow \frac { { b }^{ 2 } }{ { a }^{ 2 } } =\frac { 1 }{ 3 } \)
Now, \(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } \)
\(e=\sqrt { 1+\frac { 1 }{ 3 } } \Rightarrow \quad e=\frac { 2 }{ \sqrt { 3 } } \)

We have, foci of the ellipse at (\(\pm\)3,0) which are on the x-axis.
therefore, the equation of the ellipse is of the form
\(\Rightarrow a(\frac { 2a }{ 5 } )=2\Rightarrow { a }^{ 2 }=5\)
\(e=\frac { 2\sqrt { 5 } }{ 5 } =\frac { 2 }{ \sqrt { 5 } } and\quad { b }^{ 2 }={ a }^{ 2 }(1-{ e }^{ 2 })\)
\(\Rightarrow { b }^{ 2 }=5(1-\frac { 4 }{ 5 } )=5\times \frac { 1 }{ 5 } \)
\({ b }^{ 2 }=1\)
Hence,the equation of ellipse is \( \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 1 } =1\)
\(\Rightarrow { 17b }^{ 2 }+9={ 9b }^{ 2 }+{ b }^{ 4 }\)
\(\Rightarrow { b }^{ 4 }-{ 8b }^{ 2 }-9=0\)
\(but{ \quad b }^{ 2 }\neq -1\Rightarrow { b }^{ 2 }=1\)
From eq.(ii), we get
\({ a }^{ 2 }=9+{ b }^{ 2 }\Rightarrow { a }^{ 2 }=9+9\)
\(\Rightarrow { a }^{ 2 }=18\)
On putting the values of a² and b2 in equ(i) we get
\(\frac { { x }^{ 2 } }{ 18 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow { x }^{ 2 }+{ 2y }^{ 2 }=18\)

Given equation of circle is
x² + y² + 6x -4y + 4 = 0
(x² + 6x) + (y² - 4y) = -4
(x² + 6x + 9 - 9) + (y² - 4y + 4 - 4) = -4
(x² + 6x + 9) + (y² -4y + 4) = -4 + 4 + 9
(x + 3)² + (y - 2)² = 9
{x -(-3)}² + (y - 2)² = 3²
On computing with|(x - h)² + (y - k)² = y², we get
h=-3, k=2 and r=3
Hence, center of circle=(-3, 2)and radius=3.