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.)

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.
\(\therefore \left( \frac { k }{ 2 } \right) ^{ 2 }=\frac { 5 }{ 8 } \times 2\Rightarrow \frac { { k }^{ 2 } }{ 4 } =\frac { 5 }{ 4 } \Rightarrow k=\sqrt { 5 } =2.23m(approx.)\)

First, we draw the parabola in the positive side of x-axis and inside that, draw an equilateral \(\triangle OAB.\)
\(\sin{ 30 }^{ \circ }=\frac { PB }{ OB } \Rightarrow \frac { 1 }{ 2 } =PB\Rightarrow PB=2\)
\(and \cos{ 30 }^{ \circ }=\frac { OP }{ OB } \Rightarrow \frac { \sqrt { 3 } }{ 2 } =OP\Rightarrow OP=\frac { \sqrt { 3 } }{ 2 } \)

Coordinates of B=(OP, PB) = \(\left( \frac { \sqrt { 3 } }{ 2 } ,\frac { }{ 2 } \right) \)will satisfy
\({ y }^{ 2 }=4ax\) \(=8\sqrt { 3a } \)
Hence, the length of side of the triangle is \(=8\sqrt { 3a } \) .

Let POQ be the parabolic reflector which is 20 cm in diameter and 5 cm deep.

Then, PQ = 20 cm and OR = 5 cm, where R is the midpoint of PQ. We take OX as X-axis and OY as Y-axis. The equation of parabola may be taken as 4ax. Since, the point P(5, 10) lies on the parabola.
\({ 10 }^{ 2 }=4a(5)\Rightarrow a=5\)
Therefore, the coordinate of the focus are (a, o) i.e. (5, 0). Hence, the focus is the midpoint of the given diameter.

Let AB be the rod making an angle \(\theta\) with
OX  and P (x, y) the point on it

such that AP = 6 cm.
Since AB = 15 cm, we have
PB = 9 cm.
From P draw PQ and PR perpendiculars on y-axis and x-axis, respectively
\(\text { From } \Delta \mathrm{PBQ}, \cos \theta=\frac{x}{9} \)
\(\text { From } \Delta \mathrm{PRA}, \sin \theta=\frac{y}{6} \)
\(\text { Since } \quad \cos ^{2} \theta+\sin ^{2} \theta= 1 \)
\( \left(\frac{x}{9}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1\)
\(\text { or } \ \frac{x^{2}}{81}+\frac{y^{2}}{36}=1\)
Thus the locus of P is an ellipse

(i) Comparing the equation \(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1,\) with the standard equation
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Here, a = 3, b = 4 and c \(=\sqrt{a^{2}+b^{2}}=\sqrt{9+16}=5\)
Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 3, 0).Also, The eccentricity \(e=\frac{c}{a}=\frac{5}{3}\) The latus rectum \(=\frac{2 b^{2}}{a}=\frac{32}{3}\)
(ii) Dividing the equation by 16 on both sides, we have \(\frac{y^{2}}{16}-\frac{x^{2}}{1}=1\)
Comparing the equation with the standard equation \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) we find that a = 4, b = 1 and
\(c=\sqrt{a^{2}+b^{2}}=\sqrt{16+1}=\sqrt{17} \text { . }\)
Therefore, the coordinates of the foci are (0, ± \(\sqrt{17}\) ) and that of the vertices are (0, ± 4). Also The eccentricity \(e=\frac{c}{a}=\frac{\sqrt{17}}{4}\) The latus rectum \(=\frac{2 b^{2}}{a}=\frac{1}{2}\)

The vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ axis of parabola is y-axis.
So the parabola is of the form x² = 4ay.
Since the parabola passes through point (5,2)
∴ (5)² = 4a x 2⇒ 25=8a ⇒ a=\(\frac { 25 }{ 8 } \)
The required equation of parabola is
x²=\(\frac { 4\times 25 }{ 8 } \)y 
⇒ x²=\(\frac { 25 }{ 2 } \) y 
⇒ 2x²=25y

The given equation of parabola x²=6y
which is of the form x²=4ay
∴ 4a=6
⇒ a=\(\frac { 6 }{ 4 } \) ⇒ a=\(\frac { 3 }{ 2 } \)
∴ Coordinates  of focus are \(\left( 0,\frac { 3 }{ 2 } \right) \)
Axis of parabola is x = 0
Equation of the directrix is
y=\(\frac { -3 }{ 2 } \) ⇒  2y+3=0
Length of latus rectum =\(\frac { 4\times 3 }{ 2 } \)=6

The given equation is y² = 12x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax, we obtain
4a = 12 ⇒ a = 3
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12

The given equation of circle is
x² + y² - 8x + 10y - 12 = 0
∴ (x² - 8x) + (y² + 10y) = 12
⇒ [x² - 8x + (4)²] + [y² + 10y + (5)²]
= 12 + (4)² + (5)²
⇒ (x - 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x - 4)² + (y + 5)² = 53
⇒ (x - 4)² + (y + 5)² =(\(\sqrt { 53 } \))²
Comparing it with (x - h)2 + (y - k)2 = r2, we have
h = 4, k = - 5 and r=\(\sqrt { 53 } \)
Thus co-ordinates of the centre is (4, -5) and radius is \(\sqrt { 53 } \) .

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.
Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

\(\text { The equation of the ellipse will be of the form } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { , where } a \text { is the semi-major axis }\)
\(\text { Accordingly, } 2 a=10\)
\(a=5\)
\(\text { Distance between the foci }(2 c)=8\)
c = 4
\(\text { On using the relation } c=\sqrt{a^{2}-b^{2}} \text { , we obtain }\)
\(4=\sqrt{25-b^{2}} \)
\(\Rightarrow 16=25-b^{2} \)
\(\Rightarrow b^{2}=25-16=9 \)
\(\Rightarrow b=3\)
\(\text { Thus, the equation of the path traced by the man is } \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 \text { . }\)

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

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