F is (1, 3) and directrix is x - y + 2 = 0
x - y + 2 = 0
Let P(x,y) be any point on the parabola.
For parabola \(\frac { FP }{ PM } =1\) 
FP = PM
\({ FP }^{ 2 }=\left( x-1 \right) ^{ 2 }+\left( y-3 \right) ^{ 2 }\)
\(={ x }^{ 2 }-2x+1+{ y }^{ 2 }-6y+9\)
\( ={ x }^{ 2 }+{ y }^{ 2 }-2x-6y+10\)
\(PM=\pm \cfrac { \left( x-y+2 \right) }{ \sqrt { 1 } +1 } \)
\(=\pm \cfrac { \left( x-y+2 \right) }{ \sqrt { 2 } }\)
\( { PM }^{ 2 }=\cfrac { \left( x-y+2 \right) ^{ 2 } }{ 2 } \)
\(=\cfrac { { x }^{ 2 }+{ y }^{ 2 }+4-2xy-4y+4x }{ 2 } \)
\({ FP }^{ 2 }={ PM }^{ 2 }\)
\( { 2x }^{ 2 }+{ 2y }^{ 2 }-4x-12y+20-={ x }^{ 2 }+{ y }^{ 2 }-2xy+4x+4\) 
The required equation of the parabola is
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2xy-8x-8y+16=0\)

Solution: Let p (x₁,y₁) be any point on the locus, such that the square of its distance from A (3, -2) is equal to its distance from 5x - 12y = 13.
\(\therefore\) (x₁ - 3)² + (y₁ + 2)² = \(\frac { \left| { 5x }_{ 1 }-12{ y }_{ 1 }+13 \right| }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right) }^{ 2 } } } \)
⇒ 13[(x₁ - 3)² + (y₁ + 2)²] = 土(5x₁ - 12y₁ + 13)
⇒ 13(x₁² - 6x₁ + 9 + y₁² + 4 + 4y₁) = 士(5x₁ - 12y₁ + 13)
Case (i)
⇒ 13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = 5x₁ - 12y₁ + 13
⇒ 13x₁² + 13y₁² - 83x₁ + 64y₁ + 182 = 0
Case (ii)
13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = -(5x₁ - 12y₁ + 13)
13x₁² + 13y₁² - 73x₁ + 40y₁ + 156 = 0
\(\therefore\) Locus of (x₁, y₁) is 13x² + 13y² - 83x + 64y + 182 = 0 (or) 13x² + 13y² - 73x + 40y + 156 = 0

Let p(x,y) be any point on the parabola whose focus is F(-3, 2) and the directrix is x + y - 4 = 0.
Draw pm perpendicular to x + y - 4 = 0
Then FP = pm \(\Rightarrow\) FP² = pm²
\(\Rightarrow { (x+3) }^{ 2 }+{( y-2) }^{ 2 }={ \left[ \frac { x+y-4 }{ \sqrt { 1+1 } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }-4y+4=\frac { { x }^{ 2 }+{ y }^{ 2 }+16+2xy-8x-8y }{ 2 } \)
\(\Rightarrow\) 2(x² + y² + 6x - 4y + 13) = x² + y² + 2y - 8x - 8y + 16
\(\Rightarrow\) x² + y² - 2xy + 20x + 10 = 0.

The centre is the point of intersection of the diameters.
Solving 2x - 3y + 12 = 0 ....(1)  and
x + 4y - 5 = 0  ....(2)
(1)     \(\rightarrow\)    2x - 3y + 12 = 0
                           -      -      +
(2)\(\times\)2 \(\rightarrow\)   2x + 8y - 10 = 0
                   ___________________
                       -11y + 22 = 0
\(\Rightarrow \) -11y = -22 
\(\Rightarrow \) y = 2
Substituting y = 2 in (2) we get,
x + 4(2) - 5 = 0 
\(\Rightarrow \) x + 8 - 5 = 0 
\(\Rightarrow \) x + 3 = 0 
\(\Rightarrow \) x = -3.
\(\therefore \)  (-3, 2) is the center of the circle.
Also, given area = 154  \(\Rightarrow \) \(\pi\)r² = 154
\(\Rightarrow \frac { 22 }{ 7 } \times { r }^{ 2 }=5\)
\(\Rightarrow { r }^{ 2 }=\frac { 154\times 7 }{ 22 } =\frac { 14\times 7 }{ 2 } \)
\(\Rightarrow\) r² = 49  \(\Rightarrow\) r = 7.
\(\therefore \)  Equation of the circle is (x + 3)² + (y - 2)² = 49
\(\Rightarrow\)  x² + 6x + 9 y² - 4y + 4 = 49
\(\Rightarrow\)  x² + y² + 6x - 4y - 36 = 0.

The Condition for concurrent lines is
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=1\)
Consider = \(\left| \begin{matrix} 4 & 3 & -10 \\ 3 & -4 & 5 \\ 5 & 1 & -7 \end{matrix} \right| \)
\(\Rightarrow\) 4(28 - 5) - 3(-21 - 25) - 10(3 + 20)
\(\Rightarrow\)  92 + 138 - 230 = 0
Hence, the given lines are concurrent.

x² + y² = 16
\(\therefore\)  Centre is (0,0), r² = 16
r = 4 units

Let P (x₁,y₁) be any point on the locus and  A (-1,1) and B (2,3) are the two given points
Area Of triangle  \(\triangle\)APB  = 8sq. units
= \(\frac { 1 }{ 2 } \)[x₁ (y₂ -y₃) + x₂ (y₃ - y₁)+ x₃ + (y₁ - y₂)]
\(\Rightarrow\)  \(\frac { 1 }{ 2 } \) [x₁ ( 1 - 3) + (-1) (3 - y₁) + 2 (y₁ - 1)] = 8
x₁ ( 1 - 3) + (-1) (3 - y₁) + 2 (y₁ - 1) = 16
\(\Rightarrow\)  -2x₁ -3 +y₁ +2y₁ - 2 - 16 = 0
\(\Rightarrow\)  -2x₁ + 3y₁ + 21 = 0
Locus of P(x₁, y₁) is 2x - 3y + 21 = 0

On x axis y = 0
Let P (x₁, 0) be any point on the locus and A (7, -6) and B (3 , 4) are the two given points
PA = PB
\(\Rightarrow\) PA² = PB²
\(\Rightarrow\)  (x₁- 7)² + (0+ 6)^{2  =} (x₁ - 3)² + ( 0 - 4)²
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }\) - 14x₁ + 46 + 36    = \({ x }_{ 1 }^{ 2 }\) - 6x₁ +9 +16
\(\Rightarrow\)  -8x₁ = - 60
\(\Rightarrow\)   x₁ \( =\frac { 15 }{ 2 } \)
Required point is \(\left(\frac{15}{2}, 0\right)\)

\(4 a=1 \Rightarrow a=\frac{1}{4}\)
Equation x = a
\(x=\frac{1}{4}\)

The given parabola x²=6y is of the form x²=4 ay where 4a=6 \(\Rightarrow a=\frac { 6 }{ 4 } \Rightarrow a=\frac { 3 }{ 2 } \)
Focus is (0, a)=\(\left( 0,\frac { 3 }{ 2 } \right) \)
Vertex is (0, 0)
Equation of the directrix is y=-a \(\Rightarrow\) \(y=\frac { -3 }{ 2 } \Rightarrow 2y+3=0\)
Length of the latus section = 4a=6 units.

Given pair of lines is
3x²+10xy+8y²+14x+22y+15=0
2h=10
Here a=3, h=5, b=8,
Let \(\theta\) be the angle between the pair of lines
Then \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

Equation of circle is x² + y² = r²
r = 2
\(\Rightarrow\)  x² + y² = 4
\(\Rightarrow\)  x² + y² - 4 = 0

Equation of circle is (x-h)² + (y - k)² = r²
(h, k) = (3, 5), r = 5
(x - 3)² + (y - 5)² = 5²
\(\Rightarrow\)  x² - 6x + 9 + y² - 10y + 25  = 25 
\(\Rightarrow\)  x² + y² - 6x - 10y + 9 = 0

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)

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.