Let vertices \(\equiv \) (0, \(\pm \)b) = (0, \(\pm \)6)
   b = 6 and e = \(\frac { 5 }{ 3 } \)
\(\therefore { e }^{ 2 }=\frac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } \Rightarrow \frac { 25 }{ 9 } =\frac { 36+{ a }^{ 2 } }{ 36 } \Rightarrow { a }^{ 2 }=48\)
Hence, the required equation of hyperbola is
\(-\frac { { x }^{ 2 } }{ 48 } +\frac { { y }^{ 2 } }{ 36 } =1\Rightarrow \frac { { y }^{ 2 } }{ 36 } -\frac { { x }^{ 2 } }{ 48 } =1\)
\(\therefore foci=(0,\pm be)=(0,\pm 10)\)

Here, 2c = 16  \(\Rightarrow \) c = 8
\(e=\frac { c }{ a } \Rightarrow \sqrt { 2 } =\frac { 8 }{ a } \Rightarrow a=4\sqrt { 2 } \)
c² = a² + b² 
x² - y² = 32

\(\text{Let equation of ellipse be} \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,\ a>b\)
\(\therefore \quad \frac { 16 }{ { a }^{ 2 } } +\frac { 9 }{ { b }^{ 2 } } =1\quad and\quad \frac { 1 }{ { a }^{ 2 } } +\frac { 16 }{ { b }^{ 2 } } =1\Rightarrow { b }^{ 2 }=\frac { 247 }{ 15 } and\quad { a }^{ 2 }=\frac { 247 }{ 7 } \)
Ans. \(7{ x }^{ 2 }+15{ y }^{ 2 }=247\)

\(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 11 } =1\)

parabola y² = 4ax passes through the point (3,2).
(2)² = 4a(3)  \(\Rightarrow a=\frac { 1 }{ 3 } \)
Then, length of latusrectum = 4a= \(\frac { 4 }{ 3 } \)
Ans \({ 4 }/{ 3 }\)

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

Given circles are x²+ y²-1 = 0.....(i)
x² + y² - 2x - 6y - 6 = 0.....(ii)
and   x²+ y²- 4x 12y - 9 = 0.....(iii)
Let r₁,r₂ and r₃ be the radii of circle (i), (ii) and (iii), respectively.
We know that the general form of the circle is
x²+ y²+ 2 gx + 2 fy + c = 0
Now, comparing Eq.(i) with Eq. (iv), we get
2g = 0 \(\Rightarrow\)g = 0, 2f = 0
f = 0 and c = -1
Radius (r) = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { { 0 }^{ 2 }-{ 0 }^{ 2 }-(-1) } =1\)
On comparing Eq.(ii) with Eq.(iv), we get
2g = -2 \(\Rightarrow\)g = - 1, 2f  = - 6
f = -3 and c = - 6
Radius (r₂) = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { (-1)^{ 2 }+(-3)^{ 2 }+6 } \)
\(=\sqrt { 1+9+6 } =\sqrt { 16 } =4\)
Again, comparing Eq.(ii) with Eq.(iv), we get
2g = - 4 \(\Rightarrow\) g = -2
2f = -12
f = - 6 and c = - 9
Radius (r₂) = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { (-2)^{ 2 }+(-6)^{ 2 }-(-9) } \)
\(=\sqrt { 4+36+9 } =\sqrt { 49 } =7\)
Now, r₂-r₁ = 4 -1 =3, r₃ - r₂ = 7 - 4 =3
So, r₁,r₂ and r₃ are in arithmetic progression.
Hence proved

\(Given\quad equation\quad of\quad line\quad is\quad lx+my+n=0.\)
\(\Rightarrow y=\frac { -lx-n }{ m }\)
\(and\quad equation\quad of\quad parabola\quad is\quad { y }^{ 2 }=4ax\)
\(From\quad Eqs.\quad (i)\quad and\quad (ii),\quad we\quad get\)
\({ \left( \frac { { -lx-n } }{ m } \right) }^{ 2 } =4ax\)
\(\Rightarrow { l }^{ 2 }{ x }^{ 2 }+2lxn+{ n }^{ 2 } =4{ m }^{ 2 }ax\)
\(\Rightarrow { l }^{ 2 }{ x }^{ 2 }+2lxn-4a{ m }^{ 2 }x+{ n }^{ 2 }=0\)
\(\Rightarrow { l }^{ 2 }{ x }^{ 2 }=x(2ln-4a{ m }^{ 2 })= { n }^{ 2 }=0\)
\(Since,\quad the\quad line\quad lx+my=n\quad touches\quad the\quad parabola.\)
\(So,Eq.(iii)\quad have\quad equal\quad roots.\)
\( i.e\quad discriminant\quad (D)=0\quad \Rightarrow \quad { B }^{ 2 }-4AC=0\)
\(\Rightarrow { (2ln-4a{ m }^{ 2 }) }^{ 2 }-{ 4l }^{ 2 }{ n }^{ 2 }=0\)
\( \Rightarrow 4{ l }^{ 2 }{ n }^{ 2 }-16lna{ m }^{ 2 }=16{ a }^{ 2 }{ m }^{ 4 }-{ 4l }^{ 2 }{ n }^{ 2 }=0\)
ln = am²
Hence proved

Parabola is symmetrical about Y-axis and passes through (2, -3)
So, equation of parabola is of the form x² = -4 day
On putting x= 2, y=-3, we get
4 = -4a(-3)\(\Rightarrow a=\frac { 1 }{ 3 } \)

Given, centre (h, k) = (1, 2)
and circle touches on X-axis.
\(\therefore\) Radius (r) = y-coordinate of centre = 2
So, equation of circle is
\((x-1)^{2}+(y-2)^{2}=2^{2} \quad\left[\because(x-h)^{2}+(y-k)^{2}=r^{2}\right] \)
\(\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4=4 \)
\(\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\Rightarrow x^{2}+y^{2}-2 x-4 y+1=0\)
which is the required equation of circle.

Here, wire are vertical.
Let equation of the parabola be in the form
\({ x }^{ 2 }=4ay\)    ...(i)

Focus is at the middle of the cable and shortest and longest vertical supports are 6 m and 30 m and roadway in 100 m long.
Clearly, the coordinates of Q(50, 24) will satisfy Eq.(i)
\(\therefore \quad (50)^{ 2 }=4a\times 24\Rightarrow 2500=96a\Rightarrow a=\frac { 2500 }{ 96 } \)
\(Hence,from\quad Eq.(i),\quad { x }^{ 2 }=4\times \frac { 2500 }{ 96 } y\Rightarrow { x }^{ 2 }=\frac { 2500 }{ 24 } y\)
 Let PR=km
Then, point\ P(18,k) will satisfy the equation of parabola.
\(\therefore \quad From\quad Eq.(i),\quad (18)^{ 2 }=\frac { 2500 }{ 24 } \times k\)
\(\Rightarrow 324=\frac { 2500 }{ 24 } k\Rightarrow k=\frac { 324\times 24 }{ 2500 } =\frac { 324\times 6 }{ 625 } =\frac { 1944 }{ 625 } \)
\( \Rightarrow k=3.11\)
Therefore  Required length= 6+k=6+3.11=9.11m(approx.)

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.
Length of latusrectum,
\(\frac { { 2b }^{ 2 } }{ a } =\frac { 2\times 16 }{ 7 } =\frac { 32 }{ 7 } \)

Given equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1.\)
Since, denominator of \(\frac { { x }^{ 2 } }{ 36 } \) is greater than denominator of \(\frac { { y }^{ 2 } }{ 16 } \) .

On comparing the above equation with \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1.\)
\(we\quad get\quad a=6\quad and\quad b=4.\)
\(\therefore Major\quad axis=2a=2\times 6=12\)
\(minor\quad axis=2b=2\times 4=8\)
\(value\quad of\quad c=\sqrt { { a }^{ 2 }-{ b }^{ 2 } } =\sqrt { (6)^{ 2 }-(4)^{ 2 } } \)
\(=\sqrt { 36-16 } =\sqrt { 20 } =2\sqrt { 5 } \)
\(Vertices\quad are\quad (6,0)\quad and\quad (-6,0)\)
\(Eccentricity,e=\sqrt { 1-\left( \frac { b }{ a } \right) ^{ 2 } } =\sqrt { 1-\left( \frac { 4 }{ 6 } \right) ^{ 2 } }\)
\(=\sqrt { \frac { 36-16 }{ 36 } } =\sqrt { \frac { 20 }{ 36 } } =\frac { \sqrt { 20 } }{ 6 } \)
\(Directrices\quad are\quad x=\pm \frac { a }{ e } =\pm \frac { 6 }{ \sqrt { 20/6 } } =\pm \frac { 36 }{ \sqrt { 20 } }\)
\( and\quad length\quad of\quad latusrectum=\frac { 2{ b }^{ 2 } }{ a } =\frac { 2(4)^{ 2 } }{ 6 } =\frac { 16 }{ 3 } \)

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)
\(e=\frac { \sqrt { 13 } }{ 2 } \)

\(\sqrt { { (x-1) }^{ 2 }+{ (y+1) }^{ 2 } } =\frac { 1 }{ 2 } .\frac { \left| x-y-3 \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \)
\(\Rightarrow 8[({ x }^{ 2 }+1-2x)+({ y }^{ 2 }+1+2y)]\)
\(={ x }^{ 2 }+{ y }^{ 2 }+9-2xy+6y-6x\)
Ans. \(7{ x }^{ 2 }+7{ y }^{ 2 }+2xy+10x-10y+7=0\)