Given, \(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 20 } \) =1.
On comparing the above equation with
\(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 20 } \) =1, we get
a² = 36 \(\Rightarrow \) a = 6 and b² = 20 \(\Rightarrow \) b = 2\(\sqrt { 5 } \) 
\(\because \ e=\frac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } =\frac { \sqrt { 36-20 } }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Distance between the directrices = \(\frac { 2a }{ e } =\frac { 2X6 }{ 2/3 } \) = 18.

Given, equation of parabola is .y² = -8x, which is of the form .y² = 4ay i.e focus lies on the positive direction of x-axis
Here, 4a = 2 \(\Rightarrow\) a = 2                  
\(\therefore\)  Focus = (-a, 0) =\(\left(-2 ,0 \right) \) 
Axis = X-axis
Directrix, x = a \(\Rightarrow\) x = 2 and length of latusrectum = 4a = 4\(\times\)2 = 8

On comparing the given equation with (x - h)² + (y - k)² = r² , we get
\(h=\frac { 1 }{ 2 } ,k=-\frac { 1 }{ 3 } \)
\(and\quad r=\frac { 1 }{ 2 } \)
\((\frac { 1 }{ 2 } ,-\frac { 1 }{ 3 } )\quad and\quad \frac { 1 }{ 2 } \)

Given center is (-a,-b)
 \(\therefore h=-a,k=-b\)   and radius (r) =\(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)
On putting these values in equation of circle
\((x-h)^{ 2 }+(y-k)^{ 2 }={ r }^{ 2 }\)  we get
\([x-(-a)]^{ 2 }+[y-(-b)]^{ 2 }=(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } )^{ 2 }\)
\(\Rightarrow (x+a)^{ 2 }+(y+b)^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+a^{ 2 }+2ax+{ y }^{ 2 }+b^{ 2 }+2by={ a }^{ 2 }-{ b }^{ 2 }\quad [\because (A+B)^{ 2 }={ A }^{ 2 }+2AB+B^{ 2 }]\)
\(\Rightarrow { x }^{ 2 }+y^{ 2 }+2ax+2by+{ a }^{ 2 }+{ b }^{ 2 }=0\)
\(\therefore { x }^{ 2 }+y^{ 2 }+2ax+2by+2b^{ 2 }=0\)
Which is the required equation of circle

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
\(\frac { { x }^{ 2 } }{ 18 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow { x }^{ 2 }+{ 2y }^{ 2 }=18\)

Let B(x₁,y₁) be coordinates of the point of intersection of axis and directrix. Then A(-2,2) is the mid-point of line segment joining F(-6,-6) and B(x₁,y₁)
\(\therefore -2=\frac { x_{ 1 }-6 }{ 2 } \Rightarrow x_{ 1 }=2 \ and \ 2=\frac { { y }_{ 1 }-6 }{ 2 } \ \Rightarrow \ y_{ 1 }=10\)
So, the point B(2,10) is the point of intersection of axis and directrix.

Now, slope of line segment joining vertex and focus is
\({ m }_{ 1 }=\frac { -6-2 }{ -6+2 } =\frac { -8 }{ -4 } =2\)
\( \therefore Slope\quad of\quad directrix,\quad m_{ 2 }=\frac { -1 }{ 2 } \left[ \because AF\quad \bot \quad NB,\quad so\quad m_{ 1 }{ m }_{ 2 }=-\Rightarrow m_{ 2 }=\frac { -1 }{ m_{ 1 } } \right] \)
\(Then,\quad the\quad equation\quad of\quad directrix\quad is\quad y-10=\frac { -1 }{ 2 } (x-2)\)
\(\Rightarrow 2y-20=-x+2 \Rightarrow x+2y=22 \)
Let P(x,y) be any point on parabola and PN be the length of perpendicular from P on directrix and FP be the distance between focus F and point P
\(So,\quad FP=PN\Rightarrow (FP)^{ 2 }=(PN)^{ 2 }\)
\(\Rightarrow (x+6)^{ 2 }+(y+6)^{ 2 }=\left( \frac { x+2y-22 }{ \sqrt { 1+4 } } \right) ^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+36+12x+y^{ 2 }+36+12y\)
\(=\frac { { x }^{ 2 }+4y^{ 2 }+484+4xy-44x-88y }{ 5 } \)
\(\Rightarrow 5x^{ 2 }+180+60x+5y^{ 2 }+60y+180\)
\(={ x }^{ 2 }+4y^{ 2 }+484+4xy-44x-88y\)
\(\Rightarrow { 4x }^{ 2 }+{ y }^{ 2 }-4xy+104x+148y-124=0\)
\(which\quad is\quad the\quad required\quad equation\quad of\quad parabola\)

Given equation of ellipse is \(​​\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 16 } =1.\)
\(\text{On comparing with} \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\text{,we get }a=7,b=4\)

Here, a>b, so major axis is along X-axis.
\(Eccentricity,e=\frac { c }{ a } =\frac { \sqrt { 33 } }{ 7 } \)

Given equation of ellipse is \(​​\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 16 } =1.\)
\(\text{On comparing with} \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\text{,we get }a=7,b=4\)

Here, a > b, so major axis is along X-axis.
Major axis, 2a = 2 x 7 = 14

\(\frac { { y }^{ 2 } }{ \frac { 36 }{ 5 } } -\frac { { x }^{ 2 } }{ 4 } -=1\)
Here , \({ a }^{ 2 }=\frac { 36 }{ 5 } \)  and b² = 4  \(c=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)
\(=\sqrt { \frac { 36 }{ 5 } +4 } =\sqrt { \frac { 56 }{ 5 } } =2\sqrt { \frac { 14 }{ 5 } } \)
\(\left( 0,\pm \frac { 2\sqrt { 14 } }{ \sqrt { 5 } } \right) ,\left( 0,\pm \frac { 6 }{ \sqrt { 5 } } \right) ,\frac { \sqrt { 14 } }{ 3 } ,\frac { 4\sqrt { 5 } }{ 3 } \)

Let Since, it is passes through (3,0) and  \((3\sqrt { 2 } ,2)\)
\(\frac { 9 }{ { a }^{ 2 } } -0=1\)  and  \(\frac { 18 }{ { a }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } =1\)
a² = 9 and b² = 4
b² = a² (e²-1)
\(e=\frac { \sqrt { 13 } }{ 2 } \)