Given F₁P + F₂P = 8
By the focal property of ellipse
F₁P + F₂P = 2a
∴ 2a = 8 ⇒ a = 4
and distance between the foci = F₁F₂ = 6
2ae = 6 ⇒ ae = 3
∴ 4(e) = 3 ⇒ e \(\frac34\)

∴ b² = a²(1- e²)
= \(16\left( { 1-\left( \frac { 3 }{ 4 } \right) }^{ 2 } \right) =16\left( 1-\frac { 9 }{ 10 } \right) =16\left( \frac { 7 }{ 16 } \right) =7\)
∴ The path traced by him is an ellipse and its equation is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 7 } \) = 1

Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { y^{ 2 } }{ 9 } =1\)
∴ a² = 25, b² = 9
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
Focus is (ae, 0) = \(\left( 5\times \frac { 4 }{ 5 } \right) \) = (4, 0)
Since the focus of the hyperbola coincides with the focus of the ellipse, foci of the hyperbola are (±4,0).
Let A be the length of the semi-transverse axis
∴ Ae - 4 ⇒ 2A = \(\frac { 4 }{ e } =\frac { 4 }{ 2 } =2\) [∵ e = 2]
Let B b th length of the semi conjugate axis
B² = A²(e² - 1) = 4(4 - 1) = 12
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

Let us take the axis AX as the and the tangent AY at A as y-axis.
Equation of the parabola is y² = 4ax
CA 15, FG = 120
CF = CG= 60
F is (15, 60)
Since F lies on (1), 602 = Aa(15) ∴ ⇒ a = 60
y² = 240x
When y = 24, 24² = 240(x)
⇒ x = \(\frac { 24\times 24 }{ 240 } \)
x = \(\frac{24}{10}\) = \(\frac{12}{5}\) = 2.4
From the diagram
BD = BE - ED = 15-2.4 = 12.6m.
Hence the required height is 12.6 m.

Let the equation of the parabola be y² = 4ax
Let AB = I.

Since ABC is an equilateral triangle,
∠BAM = ∠MAC = 30^o
In MBM, cos 30°= \(\frac{AM}{l}\)
⇒ AM = l cos 30° = l\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \)
sin 30^o = \(\frac { BM }{ AM } \)
⇒ sin 30^o = \(\frac { BM }{ l } \)
⇒ BM = l sin 30^o = l\(\left( \frac { 1 }{ 2 } \right) \)
ஃ Co-ordinates of B are (AM, BM) ⇒ \(B\left( \frac { \sqrt { 3 } l }{ 2 } ,\frac { 1 }{ 2 } \right) \)
Also B lies on (1)
\({ \left( \frac { l }{ 2 } \right) }^{ 2 }=4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
⇒ \(\frac { { l }^{ 2 } }{ 4 } =4a\left( \frac { \sqrt { 3 } l }{ 2 } \right) \)
\(\frac { { l }^{ 2 } }{ 4 } =2\sqrt { 3 } al\)
\(\frac { l }{ 4 } =2\sqrt { 3 } \) a ⇒ 1 = 8 \(a\sqrt { 3 } \)
Length of the side of the equilateral triangle is 8 \(a\sqrt { 3 } \) units.

Since the axis of the parabola is vertical, it is open upward.

Its equation is x² = 4ay .....(1)
Since the base of the parabola is 5 m
VA = 2.5 m and the height of the arch is 10 m
A(2.5, 10) lies on the parabola
(2.5)² = 4a(10)
⇒ 6.25 = 40a ⇒ a = \(\frac{625}{4000}=\frac{5}{32}\)
Substituting a = \(\frac{5}{32}\) in (1) we get,
x² = 4\(\left( \frac { 5 }{ 32 } \right) \) y ⇒ x² = \(\frac58\)y
Let x be the width of the arch, when the height is 2m.
∴ (x, 2) is a point on the parabola
∴ x² = \(\frac58\)(2) = \(\frac{5}{4}\)
∴ x = \(\frac{\sqrt5}{2}\)
and 2x = \(\frac{2\sqrt5}{2}\) = \(\sqrt5\)