\(y=\frac { 1 }{ 5 } { x }^{ 2 }-6x+100\)
\(y=\frac{x^2-30 x+500}{5}\)
\(\Rightarrow\)  5y = x² - 30x + 500
\(\Rightarrow\) x² - 30x = 5y - 500
\(\Rightarrow\) (x - 15)² = 5y - 500 + 225
\(\Rightarrow\) (x - 15)² = 5y - 275
\(\Rightarrow\) (x - 15)² = 5(y - 55)
\(\Rightarrow\) X² = 5Y is a parabola
where X = x - 15, Y = y - 55.
The vertex of the parabola is (15, 55) with respect to (x, y) axis. The output and the average cost at the vertex are 15 kg and Rs. 55.

Equation of the circle is x² + y² - 2x + 6y - 15 = 0
2g = -2, 2f = 6
g = - 1, f = 3
Centre (-g, f) = (1, -3)
Let A(4, 1) and B(x, y) be the 2 ends of the diameter
Midpoint of AB \(=\left(\frac{4+x}{2}, \frac{1+y}{2}\right)=c(1,-3)\)
\(\frac{4+x}{2}=1, \ \frac{1+y}{2}=-3\)
\(4+x=2,1+y=-6\)
x = - 2, y = - 7
B(-2, -7) is the other end of the diameter

Let m₁, m₂ be the slopes of pair of straight lines.
\(m_1+m_2=\frac{-2 h}{b}, \ m_1m_2=\frac{a}{b}\)
given m₂ = 3m₁
\(m_1+3 m_1=\frac{-2 h}{b}\)
\(4 m_1=\frac{-2 h}{b} \Rightarrow m_1=\frac{-h}{2 b}\)
\(m_2=\frac{-3 h}{2 b}\)
\(m_1 m_2=\frac{a}{b}\)
\(\left(\frac{-h}{2 b}\right)\left(\frac{-3 h}{2 b}\right)=\frac{a}{b}\)
\(3 h^2=4 \mathrm{ab}\)
Hence proved

F is (-1, -2) and directrix is 4x - 3y + 2 =0
Let P (x, y) be any point on the parabola
for parabola, \(\frac { FP }{ Pm } =1\)
FP = PM
\(\Rightarrow\) FP² = PM²
\(\Rightarrow\) (x+1)² + (y+2)² =  \(\left[ \frac { 4x-3y+2 }{ \sqrt { { 4 }^{ 2 }+(-3)^{ 2 } } } \right] ^{ 2 }\)
\(\Rightarrow\) x² + 2x +1 +y² + 4y + 4 = 16x² +9y² + 4 -24xy - 12y + 16x
\(\Rightarrow\) 25x² + 50x + 25y² +50x +100y + 125  - 16x² - 9y² - 4 + 24xy +16x - 12y = 0
\(\Rightarrow\)  9x² + 16y² + 24xy + 34x + 112y + 121 = 0

8px + (2 - 3p) y + 1 = 0
\(m_1=\frac{-a}{b}=\frac{-8 p}{2-3 p}\)
px + 8y - 7 = 0
\(m_2=\frac{-p}{8}\)
Since the lines are perpendicular \(n_1 \times m_2=-1\)
\(\frac{-8 p}{2-3 p} \times \frac{-p}{8}=-1\)
\(P^2=-2+3 P\)
\(P^2-3 P+2=0\)
\((P-1)(P-2)=0\)
P = 1, 2