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
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2xy-8x-8y+16=0\)

Here \(a=2,b=2,h=\cfrac { 5 }{ 2 } ,g=\cfrac { 15 }{ 2 } ,f=9,c=k\) 
The given line represents a pair of straight lines if,
\(abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0\)
\(i.e., \ 4k+\cfrac { 675 }{ 2 } -162\cfrac { 225 }{ 2 } -\cfrac { 25 }{ 4 } k=0\)
\( \Rightarrow 16k+1350-648-450-25k=0\)
\(\Rightarrow 9k=252\therefore k=28\)

Let (x₁, y₁) be the point of contact.
\(\therefore\) Equation of tangent at (x₁,y₁) to the circle is xx₁ + yy₁ = 9
Its slope is -\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } \)
Given that the tangent is parallel to 2x + y - 3 = 0
\(\therefore\) Their slopes must be equal
\(\therefore \quad \frac { -{ x }_{ 1 } }{ { y }_{ 1 } } =\frac { -2 }{ 1 } \Rightarrow { x }_{ 1 }=2{ y }_{ 1 }\)
But \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\)
\(\Rightarrow { (2{ y }_{ 1 }) }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\Rightarrow 4{ y }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }=9\)
\(\Rightarrow 5{ y }_{ 1 }^{ 2 }=9\Rightarrow { y }_{ 1 }=\pm \frac { 3 }{ \sqrt { 5 } } \)
\(\therefore x=\frac { 6 }{ \sqrt { 5 } } \)
Hence \(\frac { { 6x }_{ 1 } }{ \sqrt { 5 } } =\frac { 3{ y }_{ 1 } }{ \sqrt { 5 } } =9\Rightarrow 6x+3y=9\sqrt { 5 } \)
\(\Rightarrow 2x+y=3\sqrt { 5 } \)

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.

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

c = 16\(\pi\)
\(\Rightarrow\)  2\(\pi\)r  = 16 \(\Rightarrow\) r = 8
∴ Equation of the circle is \((x-h)^2+(y-k)^2=r^2\)
( x + 3)² + (y + 2)² = 8²
\(\Rightarrow\) x² + 6x + 9 +y² + 4y + 4 = 64
\(\Rightarrow\) x² + y² + 6x + 4y + 13 - 64 = 0
\(\Rightarrow\) x² + y² + 6x + 4y - 51 = 0

m₁ = \(\frac { 1 }{ 2 } \), m₂ = 3
tan \(\theta\) = \(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \)  tan \(\theta\) \(=\left| \frac { \frac { 1 }{ 2 } -3 }{ 1+\frac { 1 }{ 2 } \left( 3 \right) } \right| \)
\(\Rightarrow \) tan \(\theta\) = \(\left| \frac { \frac { -5 }{ 2 } }{ \frac { 5 }{ 2 } } \right| =1\)
\(\Rightarrow \)  tan \(\theta\)  = 1 \(\Rightarrow \)  \(\theta\) = 45^°
Angle between the line is 45^°

On y axis x = 0
\(y^2=16\)

Equation of x axis y = 0, equation of y axis x = 0
\(\therefore\) combined equation xy = 0

(0, a), a = 4

\(\sqrt{4^2+5^2-16}=\sqrt{5^2}=5\)

\(a+b=0 \Rightarrow k-2=0\)

\(m=\frac{-a}{b}=\frac{-7}{5}\)

The circle \({ x }^{ 2 }+{ y }^{ 2 }+ax+by=0\) passing through (1, 2) and (1, 1)
\(\therefore \) We have 1 + 4 + a + 2b = 0 and  1 + 1 + a + b = 0
\(\Rightarrow a+2b=-5\) (1)
and a + b = –2  (2)
Solving (1) and (2),we get a = 1,b = -3

Here \({ r }^{ 2 }=25\Rightarrow r=5\) 
Parametric equations are \(x=rcos\theta ,y=rsin\theta \) 
\(\Rightarrow x=5cos\theta ,y=5sin\theta ,0\le \theta \le 2\pi \)

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 \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

Let m₁ and m₂ be the slopes of 2x - y + 3 = 0 and x + y + 2 = 0
Now m₁ = 2, m₂ = –1
Let \(\theta \) be the angle between the given lines
tan \(\theta \) =\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(tan\quad \theta =\left| \frac { 2-(-1) }{ 1+2(-1) } \right| =3\)
\(\Rightarrow \theta ={ tan }^{ -1 }(3)\)

Equation of parabola is y² = kx
It passes through  (4, -2)
\(\Rightarrow\) 4 = 4k
\(\Rightarrow\) k = 1
\(\therefore\) Equation of the parabola if   y² = x
Now latus rectum = \(\boxed{4a = 1}\)
\(\Rightarrow\)  a = \(\frac { 1 }{ 4 } \)
Length of latus rectum = 4a = 1
\(\therefore\) Focus f = (a. 0) \(\boxed{F\left( \frac { 1 }{ 4 } ,0 \right) }\)

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