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

Here, axis is vertical, so let arch of parabola is in the form
\({ x }^{ 2 }=4ay\)  ..(i)
\(\text {Given, OB=10m}\)
\(\text {and AC}=5m\Rightarrow AB=\frac { 5 }{ 2 } m\)

\(\text{Hence,coordinates of A}=\left( \frac { 5 }{ 2 } ,10 \right) \text{will satisfy Eq.(i)}\)
\( \left( \frac { 5 }{ 2 } \right) ^{ 2 }=4a\times 10\Rightarrow \frac { 25 }{ 4 } =40a\Rightarrow a=\frac { 5 }{ 32 } \)
From Eq.(i),
\( { x }^{ 2 }=4\times \frac { 5 }{ 32 } y\Rightarrow { x }^{ 2 }=\frac { 5 }{ 8 } y\)
 Now,let OR=2 and
\( PQ=k\Rightarrow RP=\frac { k }{ 2 } \)
\(\therefore,P=\left( \frac { k }{ 2 } ,2 \right) \) will lie on parabola.
\(\therefore \left( \frac { k }{ 2 } \right) ^{ 2 }=\frac { 5 }{ 8 } \times 2\Rightarrow \frac { { k }^{ 2 } }{ 4 } =\frac { 5 }{ 4 } \Rightarrow k=\sqrt { 5 } =2.23m(approx.)\)

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

Since the axes of the parabola is a line perpendicular to the directrix and passing through the vertex (2,1).

So the equation of lines is x+y+λ=0.
∴ 2+1+λ=0⇒λ=−3
∴ equation of directrix is x+y−3=0....(i)
equation of directrix is x−y+1=0....(ii)
Solving (i) and (ii),we have
x=1 and y=2
So the co−ordinate of B are(1,2)
Now A is mid point of  BS
\( \therefore \frac { { x }_{ 1 }+1 }{ 2 } =2\quad and\frac { { y }_{ 1 }+2 }{ 2 } =1\)
\(\Rightarrow { x }_{ 1 }=3\quad and\quad { y }_{ 1 }=0\)
\(So\quad co-ordinates\quad of\quad focus\quad are\quad (3,0)\)
\(Now\quad PS=PM\Rightarrow { PS }^{ 2 }={ PM }^{ 2 }\)
\(\Rightarrow { \left( x-3 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left[ \frac { x-y+1 }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ -1 }^{ 2 } } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+9-6x+{ y }^{ 2 }\)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }+1-2xy+2x-2y }{ 2 }\)
\(\Rightarrow 2({ x }^{ 2 }+{ y }^{ 2 }-6x+9)={ x }^{ 2 }{ y }^{ 2 }+1-2xy+2x-2y\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-14x+2y+2xy+17=0.\)

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the circle passes through (0, 0),
(0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)² + (0 – k)² = h² + k² … (1)
(0 – h)² + (b – k)² = h² + k² … (2)
From equation (1), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
\(\text { However, } a \neq 0 ; \text { hence, }(a-2 h)=0 \Rightarrow h=\frac{a}{2} \text { . }\)
\(\text { From equation (2), we abtain }\)
\(h^{2}+b^{2}-2 b k+k^{2}=h^{2}+k^{2}\)
\(\Rightarrow b^{2}-2 b k=0\)
\(\Rightarrow b(b-2 k)=0\)
\(\Rightarrow b=0 \text { or }(b-2 k)=0\)
\(\text { However, } b \neq 0 \text { ; hence, }(b-2 k)=0 \Rightarrow k=\frac{b}{2} \text { . }\)


Thus, the equation of the required circle is
\(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2} \)
\(\Rightarrow\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4} \)
\(\Rightarrow 4 x^{2}-4 a x+a^{2}+4 y^{2}-4 b y+b^{2}=a^{2}+b^{2} \)
\(\Rightarrow 4 x^{2}+4 y^{2}-4 a x-4 b y=0 \)
⇒ x² + y²- ax - by = 0
which is required equation of circle.

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