Since the axis of the parabola is vertical, its equation will be x² = 4ay.
Arch is 10m high and 5 m wide at base.
​\(\therefore\)​ Point \((\frac{5}{2},10)\) lies on the parabola
​\(\therefore\)​  \(\frac{25}{4}\) = 4a(10)⇒ 4a = \(\frac{25}{40}\) = \(\frac{5}{8}\)(y)
\(\therefore\) Equation of the parabola becomes x² = \(\frac{5}{8}\)(y)
Let the width of the arch 2m from the vertex is 2b, then point (b, 2) lies on the parabola
\(\therefore\) b² = \(\frac { 5(2) }{ 8 } =\frac { 10 }{ 8 } =\frac { 5 }{ 4 } \Rightarrow b=\frac { \sqrt { 5 } }{ 2 } \)

\(\therefore\) width of arch is  2b = 2.\(\frac{\sqrt{5}}{2}\) = √5m = 2.23m(app)

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

Since the parabola is symmetric about X-axis and has its vertex at (0,0), its equation will be of the form y² = 4ax or y² = -4ax.
But the parabola passes through (2, 3) which is in the I quadrant, its equation will be of the form y² = 4ax, which is open rightward.
Substituting (2,3) in y² = 4ax, we get
9 = 4a(2) ⇒ 8a = 9
⇒ a = \(\frac{9}{8}\)
\(\therefore\) Equation of the parabola is y² = 4(\(\frac{9}{8}\))x
⇒ y² = \(\frac{9}{2}\)x
⇒ 2y² = 9x ⇒ 2y² - 9x = 0

Given equation of the circle is x² + y² = 169...(1)
Equation of the tangent at (x₁, y₁) to circle (1) is xx₁ + yy₁ = 169
Now, Equation of the tangent at (5,12) to circle (1) is
x(5) + y(12) = 169 ⇒ 5x + 12y - 169 = 0...(2)
and equation of the tangent at (12,-5) to circle (1) is
x(12) + y(-5) = 169 ⇒ 12x - 5y = 169 = 0...(3)
Let m₁ and m₂ be the slopes of the tangents (2) and (3)
ஃ m₁=\(\frac { -Co-efficient\quad of\quad x }{ Co-efficient\quad of\quad y } =\frac { -5 }{ 12 } \)
Similarly m₂ = \(\frac { -12 }{ -5 } =\frac { 12 }{ 5 } \)
Consider m₁m₂ = \(\left( \frac { -5 }{ 12 } \right) \left( \frac { 12 }{ 5 } \right) =-1\)
Since m₁m₂ = -1, the tangents at (5,12) and (12,-5) to the circle x² + y² = 169 are perpendicular to each other.

Let x represent the number of units and y its cost.
By the given data,
x₁(6000) y₁(33,000)
x₂(8000) y₂(40,000)
Using two point form, \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
⇒ \(\frac { y-33000 }{ 40,000-33000 } =\frac { x-6000 }{ 8000-6000 } \)
⇒ \(\frac { y-33000 }{ 7000 } =\frac { x-6000 }{ 2000 } \)
⇒ \(\frac { y-33000 }{ 7 } =\frac { x-6000 }{ 2 } \)
⇒ 2y - 66000 = 7x - 42000
⇒ 2y = 7x - 42000 + 66000
2y = 7x + 24000, which is the required linear function.