Centre C(–1, 3)
It lies on ax + 2y + 2 = 0
-a + 6 + 2 = 0
a = 8

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

3x² - 5xy - 2y² + 17x + y + 10 = 0
Comparing this with ax²+ 2hxy + by²+ 2gx + 2fy + c = 0
We get a=3, 2h = -5, b = -2, 
\(h=\frac { -5 }{ 2 }\)
Let \(\theta\) be the angle between the pair of straight lines then
tan \(\theta\) \(=\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
\(\Rightarrow tan\quad \theta =\frac { \pm 2\sqrt { \frac { 25 }{ 4 } +6 } }{ 1 } \)
\(\theta=\tan ^{-1}\left[\not 2\left(\frac{7}{\not 2}\right)\right]=\tan ^{-1}\)

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

\(r=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3\)

(parallel line)

x + y = 6 .....(1)  
x + 2y = 4  .....(2)
On solving (1) and (2) we get the centre of the circle
C(-h, - k) = (8, - 2)
Equation of circle is \((x-h)^2+(y-k)^2=r^2\)
\((x-8)^2+(y+2)^2=r^2\)
It passes through (2, 6)
\((2-8)^2+(6+2)^2=r^2\)
\(36+64=r^2 \Rightarrow r^2=100\)
Required equation is \((x-8)^2+(y+2)^2=100\)
\(x^2-16 x+64+y^2+4 y+4=100\)
\(x^2+y^2-16 x+4 y-32=0\)

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

Compare the equation 
\({ 2x }^{ 2 }+7xy+{ 3y }^{ 2 }+5x+5y+2=0\)  with 
\({ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0\) .we get,
\(a=2,b=3,h=\cfrac { 7 }{ 2 } ,g=\cfrac { 5 }{ 2 } ,f=\cfrac { 5 }{ 2 } ,c=2\) 
Now \(\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \right| =\left| \begin{matrix} 2 & \frac { 7 }{ 2 } & \frac { 5 }{ 2 } \\ \frac { 7 }{ 2 } & 3 & \frac { 5 }{ 2 } \\ \frac { 5 }{ 2 } & \frac { 5 }{ 2 } & 2 \end{matrix} \right| \) 
= \(2\left( 6-\cfrac { 25 }{ 4 } \right) -\cfrac { 7 }{ 2 } \left( 7-\cfrac { 25 }{ 4 } \right) +\cfrac { 5 }{ 2 } \left( \cfrac { 35 }{ 4 } -\cfrac { 15 }{ 2 } \right) \) 
\(\\ =2\left( -\cfrac { 1 }{ 4 } \right) -\cfrac { 7 }{ 2 } .\cfrac { 3 }{ 4 } +\cfrac { 5 }{ 2 } .\cfrac { 5 }{ 4 } \)
\(=-\cfrac { 1 }{ 2 } -\cfrac { 21 }{ 8 } +\cfrac { 25 }{ 8 } =0\) 
Hence the given equation represents a pair of straight lines. 
Now consider,
\({ 2x }^{ 2 }+7xy+{ 3y }^{ 2 }={ 2x }^{ 2 }+6xy+xy+{ 3y }^{ 2 }\)
\(=2x\left( x+3y \right) +y\left( x+3y \right) \)
\(=\left( x+3y \right) \left( 2x+y \right) \)
Let \({ 2x }^{ 2 }+7xy+{ 3y }^{ 2 }+5x+5y+2=\left( x+3y+1 \right) \left( 2x+y+m \right) \) 
Comparing the coefficient of x 2l + m = 5 ...(1)
Comparing the coefficient of y, l + 3m = 5 ...(2)
Solving (1) and (2), we get m = 1 and l = 2
\(\therefore \) The separate equations are
x + 3y + 2 = 0 and 2x + y + 1 = 0.

Equation of straight line is 3x + 2y + 7 = 0
At P(-1, -2) PT = -3 - 4 + 7 = 0
At Q(1, 0) QT = 3 + 0 + 7 = 10 > 0
At R (-3, -4) RT = -9 - 8 + 7 = - 10 < 0
P lies on the straight line, Q above the line and R below the line