The equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
\(\therefore \ { a }^{ 2 }=25\ and \ { b }^{ 2 }=9\)
\(\text { Eccentricity of ellipse }=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\text { So the co-ordinates of foci are } (\pm 4,0)\)
\(\text { Let equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ { a' }^{ 2 } } -\frac { y^{ 2 } }{ { b }'^{ 2 } } =1\)
\(\text { Let e' be the eccentricity of required hyperbola then,}\)
\(a'e'=4\Rightarrow 2e'=4\Rightarrow e'=2\)
\(\therefore b'^{ 2 }={ a' }^{ 2 }({ e' }^{ 2 }-1)\Rightarrow b'^{ 2 }=4(4-1)=12\)
\(\text { Thus equation of required hyperbola is}\)
\(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Let the sides AB, BC, and CA of ABC be represented by the equation x+y=2, 3x-4y=6 and x-y=0, respectively

on solving x + y = 2 and 3x - 4y = 6, we get B(2,0)
On solving 3x - 4y = 6 and x-y=0, we get  C(-6,-6)
On solving x+y=2 and x-y=0, we get A(1,1)
Let the required equation of the circle be
x² + y² + 2gx + 2fy + c = 0  ...(i)
Since, it passes through A(1,1), B(2,0) and (-6,-6), so each of these points satisfy Eq(i)
1² + 2² + 2g + 2f + c = 0
\(\Rightarrow \) 2g + 2f + c = -2  ...(ii)
2² + 0² + 4g + c = 0
\(\Rightarrow \) 4g + c + 4 = 0...(iii)
ans (-6)² + (-6)² - 12g - 12f + c = 0
\(\Rightarrow \)  12g + 12f - c = 72...(iv)
On subtracting Eqs. (ii) from (iii), we get
2g - 2f + 20   g -f = -1..(v)
On adding Eq. (iv) and Eq(iii), we get
16g+12f-68=0  4g + 3f = 17....(vi)
On solving Eqs (v) and (vi), we get g=2 and f=3
On putting g=2 in Eq(iii), we get c=-12
Hence, the required equation of circle is
x² + y² + 4x + 6y -12 = 0

Let equation of ellipse be
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,a>b\)
Since, latusrectum LL' subtrends an angle of 60 to the centre.

\(\therefore \quad \angle LOL'=90^{ \circ }\)
\(Since,\quad OM\quad is\quad the\quad median\quad of\quad \triangle OLL'.\)
\(\therefore \quad \angle LOM=\angle L'OM=45^{ \circ }\)
\(In\quad the\quad right\quad angled\quad \triangle OML,\)
\(tan{ 45 }^{ ^{ \circ } }=\frac { { b }^{ 2 }/a }{ ae } \Rightarrow { e }^{ 2 }=1-\frac { { a }^{ 2 }e }{ { a }^{ 2 } } \)
\(\Rightarrow { e }^{ 2 }+e-1=0\)
\(\therefore e=\frac { -1+\sqrt { 5 } }{ 2 } \)

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

\({ (x-2) }^{ 2 }+{ (y-2) }^{ 2 }=\frac { { 2 }^{ 2 }{ (x+y-9) }^{ 2 } }{ { 1 }^{ 2 }{ +1 }^{ 2 } } \)
\(\Rightarrow \)  x² + y² - 4x - 4y + 8 = 2(x² + y² + 81 + 2xy -18x)
x² + 4xy + y² - 32x - 32y + 154 = 0

Let F₁ and F₂ be two points where the flag parts are fixed on the ground. The origin O is the mid point of F₁F₂

∴ OF₁=OF₂=\(\frac { 1 }{ 2 } \)F₁F₂=\(\frac { 1 }{ 2 } \) x 8 =4m
∴ Coordinates of F; are (-4, 0) and F₂ are (4, 0)
Let P(α, β) be any point on the track.
∴ PF₁+PF₂ = 0
∴ \(\sqrt { (\alpha +4)^{ 2 }+(\beta -0)^{ 2 } } +\sqrt { (\alpha -4)^{ 2 }+(\beta -0)^{ 2 } } \)
=10
⇒ \(\sqrt { \alpha ^{ 2 }+16+8\alpha +{ \beta }^{ 2 } } \)
=10-\(\sqrt { { \alpha }^{ 2 }+16-8\alpha +{ \beta }^{ 2 } } \)
Squaring both sides, we have
\(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }+8\alpha +16=100+{ \alpha }^{ 2 }+{ \beta }^{ 2 }-8\alpha \)+16
=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
⇒ 16\(\alpha \)-100=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
Squaring both sides again, we have
(16\(\alpha \)-100)2=(-20\(\sqrt { { (\alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16)^{ 2 } } \)
⇒ 256\(\alpha \)²+1000-3200\(\alpha \) = 400(\(\alpha \)²+\(\beta \)²+8\(\alpha \)+16)
⇒ 256\(\alpha \)²+10000-3200\(\alpha \)=400\(\alpha \)²+400\(\beta \)²-3200\(\alpha \)+6400
⇒ 144\(\alpha \)²+400\(\beta \)²=3600
⇒ \(\frac { 144\alpha ^{ 2 } }{ 3600 } +\frac { 400{ \beta }^{ 2 } }{ 3600 } \)=1 ⇒ \(\frac { { \alpha }^{ 2 } }{ 25 } +\frac { { \beta }^{ 2 } }{ 9 } \)=1
Thus required equation of locus of point P is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } \)=1

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through point (2, 3).
\(\therefore(2-h)^{2}+3^{2}=25 \)
\(\Rightarrow(2-h)^{2}=25-9 \)
\(\Rightarrow(2-h)^{2}=16 \)
\(\Rightarrow 2-h=\pm \sqrt{16}=\pm 4 \)
\(\text { If } 2-h=4, \text { then } h=-2 . \)
\(\text { If } 2-h=-4, \text { then } h=6 .\)
Equation of required circle is
(x - 6)² + (y - 0)² = (5)²
⇒ x² + 36 - 12x + y² = 25
⇒ x² + y² - 12x + 11 = 0
When h=-2
Equation of required circle is
(x + 2)² + (y - 0)² = (5)²
⇒  x² + 4 + 4x + y² = 25
⇒  x² + y² + 4x - 21 = 0

The equation of the circle is
(x - h)² + (y - k)² = r².....(i)
Since the circle passes through point (2, 3)
∴ (2 - h)² + (3 - k)² = r²
⇒ 4 + h² - 4h + 9 + k² - 6k = r²
⇒ h² + k² - 4h - 6k + 13 = r² ...(ii)
Also the circle passes through point (-1, 1)
∴ (-1-h)² + (1- k)² = r²
⇒ 1 + h² + 2h + 1 + k² - 2k = r²
⇒ h² + k² + 2h - 2k + 2 = r²...(iii)
From (ii) and (iii), we have
h² + k² - 4h - 6k + 13 = h² + k² + 2h - 2k + 2
⇒ -6h-4k=-11
⇒ 6h + 4k = 11...(iv)
Since the centre (h, k) of the circle lies on the line x - 3y - 11 = 0
∴ h- 3k-11 = 0
⇒ h - 3k = 11...(v)
Solving (iv) and (v), we have
h=\(\frac { 7 }{ 2 } \) and k=\(\frac { -5 }{ 2 } \)
Putting these values of hand k in (ii), we have
\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\frac { -5 }{ 2 } \)=r²
⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } \)-14+15+13=r²
⇒ r²=\(\frac { 65 }{ 2 } \)
Thus equation of required circle is
\(\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)
⇒ x²+\(\frac { 49 }{ 4 } \)-7x+y²+\(\frac { 25 }{ 4 } \)+5y=\(\frac { 65 }{ 2 } \)
⇒ 4x² +49-28x+4y² +25+20y= 130
⇒ 4x² + 4y² - 28x + 20y - 56 = 0
⇒ 4(x² + y² - 7x + 5y - 14) = 0
⇒ x² + y² - 7x + 5y - 14 =0.

The equation of the circle is
(x - h)² + (y - k)² = r²... (i)
Since the circle passes through point (4, 1)
∴ (4 - h)² + (1 - k)² = r²
⇒ 16 + h² - 8h + 1 + k² - 2k = r²
⇒ h² + k² - 8h - 2k + 17 = r²...(ii)
Also the circle passes through point (6, 5)
∴ (6 - h)² + (5 - k)² = r²
⇒ 36 + h² - 12h + 25 + k² - 10k = r²
⇒ h² + k² - 12h - 10k + 61 = r²...(iii)
From (ii)and (iii), we have
h²+ k²-8h-2k+ 17=h²+ k²-12h-10k+61
⇒ 4h+ 8k= 44
⇒ h + 2k = 11....(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴  4h + k = 16      ...(v)
Solving (iv) and (v), we have h = 3 and k = 4
Putting value of hand k in (ii), we have
(3)² + (4)² - 8 x 3 - 2 x 4 + 17 = r²
∴  r²=10
Thus equation of required circle is
⇒ (x - 3)² + (y - 4)² = 10
⇒ x² + 9 - 6x + y² + 16 - 8y = 10
⇒ x² + y² - 6x - 8y + 15 =0.

Let AB be a rod of length 12 cm and P(x, y) be any point on the rod such that PA = 3 cm and PB = 9 cm.

Let AR = a and BQ = b
Then ΔARP ~ ΔPQB
∴ \(\frac { AR }{ PQ } =\frac { AP }{ PB } \)
∴ \(\frac { a }{ x } =\frac { 3 }{ 9 } \) ⇒ 9a=3a  ⇒ a=\(\frac { x }{ 3 } \)
and \(\frac { BQ }{ BP } =\frac { PR }{ PA } \)
∴  \(\frac { b }{ 9 } =\frac { y }{ 3 } \) ⇒ 3b=9y ⇒ b=3y
Now OR + AR = x + a = x+\(\frac { x }{ 3 } =\frac { 4x }{ 3 } \)
OB = OQ + BQ = y + b = y + 3y = 4y
In right angled ΔAOB
AB²= OA²+ OB²
∴ (12)² = \(\left( \frac { 4x }{ 3 } \right) ^{ 2 }\)+(4y)²
⇒ 144=\(\frac { 16{ x }^{ 2 } }{ 9 } \)+16y²
⇒ \(\frac { 16{ x }^{ 2 } }{ 9\times 144 } +\frac { 16{ y }^{ 2 } }{ 144 } \) =1 ⇒ \(\frac { { x }^{ 2 } }{ 81 } +\frac { { y }^{ 2 } }{ 9 } \)=1
which is required locus of point P and which represents an ellipse.

Here foci are (0, ±\(\sqrt10\)) which lie on y-axis.
So the equation of hyperbola in standard form is \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ foci (0, ± c) is (0, ± \(\sqrt10\))  a = \(\sqrt10\)
We know that c² = a² + b²
∴ (\(\sqrt10\))² = a² + b² ⇒ b² = 10 - a²
Since the hyperbola passes through (2, 3)
∴ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } \)=1
⇒ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ 10-{ a }^{ 2 } } \)=1
⇒ \(\frac { 9(10-{ a }^{ 2 })-4{ a }^{ 2 }={ a }^{ 2 }(10-{ a }^{ 2 }) }{ { a }^{ 2 }(10-{ a }^{ 2 }) } \)
⇒ a⁴ - 23a² + 90 = 0
⇒ a⁴ - 18a² - 5a² + 90 = 0
⇒ (a² - 18) (a² - 5) = 0
⇒ a² = 18 or a² = 5
When a²= 18 then b²= 10-18=-8 (which is not possible)
When a² = 5 then b² = 10 - 5 = 5
Thus required equation of hyperbola is
\(\frac { { y }^{ 2 } }{ 5 } -\frac { { x }^{ 2 } }{ 5 } \)=1