12x²-10xy+2y²+14x-5y+c=0
ax² +2hxy+ by² +2gx+2fy+c = 0
\(\left| \begin{matrix} 2h=-10 \\ h=-5 \end{matrix} \right| b=2\left| \begin{matrix} 2g=14 \\ g=7 \end{matrix} \right| \left| \begin{matrix} 2f=-5 \\ f=\frac { -5 }{ 2 } \end{matrix} \right| \)
af² +bg² +ch² -2fgh-abc=0 is the condition
12\(\left( \frac { 25 }{ 4 } \right) \)+2(7)²+c(-5)²-2\(\left( \frac { -5 }{ 2 } \right) \)(7)(-5)-12(2)(c)=0
75+98+25c-175-24c=0, c=2
The equation is 12x²-10xy+2y²+14x-5y+2=0
12x²-10xy+2y=(3x-y)(4x-2y)
Let 12x²-10xy+2y²+14x-5y+2=(3x-y+l)(4x-2y+m)
So that 4l+3m=14, -2l-m=-5
On solving we get t=\(\frac { 1 }{ 2 } \), m=4
∴ The separate equations are 3x-y+\(\frac { 1 }{ 2 } \)=0 ⇒ 6x-2y+1=0 and
4x-2y+4=0 ⇒ 2x-y+2=0
m₁=\(\frac { -6 }{ -2 } \)=3; m₂=\(\frac { -2 }{ -1 } \)=2
∴ tanθ=\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 3-2 }{ 1+3.2 } \right| =\frac { 1 }{ 7 } \)
θ=tan^-1\(\left( \frac { 1 }{ 7 } \right) \).

Let P (h, k) be the foot of the perpendicular from the origin O on the line \(\frac{x}{a}+\frac{y}{b}=1\) which cuts the coordinates axes at A(a, 0) and B(0, b). Then,
Slope of OP x Slope of AB = -1
\(\Rightarrow \frac{k-0}{h-0}\times \frac{b-0}{0-a}=-1\)
\(\Rightarrow\) bk = ah

\(\Rightarrow b=\frac{ah}{k}\) ......(i)
Also p(h, k) lies on \(\frac{x}{a}+\frac{y}{b}=1\)
\(\therefore \frac{h}{a}+\frac{k}{b}=1\) .....(ii)
\(\Rightarrow \frac{h}{a}+\frac{k^2}{ah}=1\)    [Using (i)]
\(\Rightarrow a=\frac{h^2+k^2}{h}\)
Substituting this values of a in (i), we obtain
\(b=\frac{h^2+k^2}{k}\)
Substituting this values of a and b in \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2},\) we obtain
\(\frac{h^2}{(h^2+k^2)^2}+\frac{k^2}{(h^2+k^2)^2}=\frac{1}{c^2}\) or, h² + k² = c²
Hence, the locus of (h, k) is x² + y² = c²

The given equation is x cos \(\alpha\) + y sin \(\alpha\) = p or \(\frac{x}{p/\cos\alpha}+\frac{y}{p/sin\alpha}=1\) ........(i)
This cuts the coordinate axes at \(A(p/\cos\alpha,0)\) and \(B(0,p/\sin\alpha)\). Let P (h, k)be the mid point of the intercept AB. Then,
\(h=\frac{p/\cos\alpha+0}{2},k=\frac{0+p/\sin\alpha}{2}\)
\(\Rightarrow h=\frac{p}{2\cos\alpha},k=\frac{p}{2\sin\alpha}\)
\(\Rightarrow \cos\alpha=\frac{p}{2h},\sin\alpha=\frac{p}{2k}\)
Here, u is a variable. to find the locus of P (h, k), we have to eliminate c.
From (i), we obtain
\(\cos^2\alpha+\sin^2\alpha=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}\)
Hence, the locus of (h, k) is \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}\)

Given equation of the line is \(x\cos\theta+y\sin\theta=p\) .
Let C (h, k) be the mid point of the given line AB where it meets the two axis atA (a, 0) andB (0, b).
Since (a, 0) lies on eq (i) then \(a\cos\theta+0=p^n\)
\(\Rightarrow a=\frac{p}{\cos\theta}\) .....(ii)
B (0, b) also lies on the eq (i) then 0 + \(b\sin\theta\) = p
\(\Rightarrow b=\frac{p}{\sin\theta}\)  ...(iii)
Since C(h, k) is the mid point of AB
\(\therefore h=\frac{0+a}{2}\Rightarrow a=2h\) and \(k=\frac{b+0}{2}\Rightarrow b=2k\)
Putting the values of a and b is eq (ii) and (iii) we get
\(2h=\frac{p}{\cos\theta}\Rightarrow\cos\theta=\frac{p}{2k}\) .....(iv)
and \(2k=\frac{p}{\sin\theta}\Rightarrow\sin\theta=\frac{p}{2k}\) ....(v)
Squaring and adding eq (iv) and (v) we get
\(\Rightarrow \cos^2\theta+\sin^2\theta=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\)
So, the locus of the mid point is \(1=\frac{p^2}{4x^2}+\frac{p^2}{4y^2}\)
\(\Rightarrow \)  4x²y² = p² (x² + y²)
Hence, the value of the filter is 4x²y² = p² (x² + y²).

The given equation of line is 5x - 12y = 3 and the given point is (3, -2).
Let (a, b) be any moving point.
\(\therefore\) Distance between (a, b) and the point (3, -2) = \(\sqrt{(a-3)^2+(b+2)^2}\) and the distance of (a, b) from the line 5x - 12y = 3 = \(|\frac{5a-12b-3}{\sqrt{25+144}}|=|\frac{5a-12b-3}{13}|\)
According to the question, we have \([\sqrt{(a-3)^2+(b+2)^2}]=|\frac{5a-12b-3}{13}|\)
\(\Rightarrow (a-3)^2(b+2)^2=\frac{5a-12b-3}{13}\)
Taking numerical values only, we have \((a-3)^2(b+2)^2=\frac{5a-12b-3}{13}\)
\(\Rightarrow a^2-6a+9+b^2+4b+4=\frac{5a-12b-3}{13}\)
\(\Rightarrow a^2+b^2-6a+4b+13=\frac{5a-12b-3}{13}\)
\(\Rightarrow\) 13a² + 13b² - 78a + 52b + 169 = 5a - 12b - 3
\(\Rightarrow\) 13a² + 13b² - 83a + 64b + 172 = 0
So, only locus of the point is 13x² + 13y² - 83x + 64y + 172 = 0
Hence, the value of the filter is 13x² + 13y² - 83x + 64y + 172 = 0.