Angle between the pair of straight lines, Condition for parallel and perpendicular lines Questions in English

(B) The given equation of the pair of lines is $2x^2 + 3xy + y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 2$,$2h = 3$ (so $h = 3/2$),and $b = 1$.
Let $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$ be the slopes of the lines.
For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the sum of slopes is $m_1 + m_2 = -2h/b = -3/1 = -3$ and the product of slopes is $m_1 m_2 = a/b = 2/1 = 2$.
The angle $\theta$ between the lines is given by $\tan \theta = |\tan(\theta_1 - \theta_2)| = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
We know that $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$.
Substituting the values: $(m_1 - m_2)^2 = (-3)^2 - 4(2) = 9 - 8 = 1$.
Thus,$|m_1 - m_2| = \sqrt{1} = 1$.
Therefore,$|\tan(\theta_1 - \theta_2)| = |\frac{1}{1 + 2}| = 1/3$.

(C) Given equation: $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$
Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$
Rearranging terms: $x^2 \sin^2 \alpha + y^2 \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$
Simplifying: $x^2 \sin^2 \alpha = x^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha$
$x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$
$-x^2 \cos(2\alpha) + xy \sin(2\alpha) = 0$
This is a homogeneous equation of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = -\cos(2\alpha)$,$H = \frac{1}{2} \sin(2\alpha)$,and $B = 0$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} \right|$.
Substituting values: $\tan \theta = \left| \frac{2\sqrt{\frac{1}{4} \sin^2(2\alpha) - 0}}{-\cos(2\alpha) + 0} \right| = \left| \frac{\sin(2\alpha)}{-\cos(2\alpha)} \right| = |-\tan(2\alpha)| = |\tan(2\alpha)|$.
Thus,$\theta = 2\alpha$.

(B) The equation of the pair of angle bisectors of the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Since the coordinate axes are the bisectors,their equations are $x = 0$ and $y = 0$,which implies their combined equation is $xy = 0$.
Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we see that the coefficient of $x^2$ and $y^2$ in the bisector equation must be zero.
This implies $a - b = 0$ is not possible as per the given condition $a \neq b$,or rather,for the bisectors to be the axes,the $x^2$ and $y^2$ terms must vanish in the bisector equation.
Actually,the condition for the coordinate axes to be the bisectors is $a + b = 0$ and $h \neq 0$ is not required,but rather $h$ must be such that the equation reduces to $xy = 0$.
Given the standard form,if $a + b = 0$,the bisectors are $x^2 - y^2 = 0$,which are $y = \pm x$. If the axes are the bisectors,then $h$ must be $0$ and $a+b=0$ is not the condition; rather,the condition is $a+b=0$ is for perpendicular lines. For axes to be bisectors,$h=0$ is the condition.

(D) The given equation is $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta - \sin \theta$,$H = \cos \theta$,and $B = \cos \theta + \sin \theta$.
The acute angle $\alpha$ between the pair of straight lines is given by $\tan \alpha = \left| \frac{2 \sqrt{H^2 - AB}}{A + B} \right|$.
Substituting the values:
$H^2 - AB = \cos^2 \theta - (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - (\cos^2 \theta - \sin^2 \theta) = \sin^2 \theta$.
$A + B = (\cos \theta - \sin \theta) + (\cos \theta + \sin \theta) = 2 \cos \theta$.
Thus,$\tan \alpha = \left| \frac{2 \sqrt{\sin^2 \theta}}{2 \cos \theta} \right| = \left| \frac{2 \sin \theta}{2 \cos \theta} \right| = |\tan \theta|$.
Since $2 \theta$ is acute,$\theta$ is also acute,so $\alpha = \theta$.

(D) The given equation is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $b(\frac{y}{x})^2+2h(\frac{y}{x})+a=0$. Let $m = \frac{y}{x}$,then $bm^2+2hm+a=0$. The roots of this quadratic equation are the slopes $m_1$ and $m_2$ of the lines. Given $h^2=ab$,the discriminant $D = (2h)^2 - 4ab = 4h^2 - 4ab = 4(h^2-ab) = 0$. Since the discriminant is zero,the roots are equal,i.e.,$m_1 = m_2$. Therefore,the ratio of the slopes is $m_1:m_2 = 1:1$.

(B) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
For this to represent a pair of lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing $(2p-3)x^2 + 2pxy - y^2 = 0$ with the standard form,we have $a = 2p-3$,$h = p$,$b = -1$,$g = 0$,$f = 0$,and $c = 0$.
Substituting these values into the condition: $(2p-3)(0)(-1) + 2(0)(0)(p) - (2p-3)(0)^2 - (-1)(0)^2 - (0)(p)^2 = 0$.
This simplifies to $0 = 0$,which is always true for any $p \in R$.
For the lines to be distinct,the condition $h^2 - ab > 0$ must be satisfied.
Here,$h^2 - ab = p^2 - (2p-3)(-1) = p^2 + 2p - 3$.
We require $p^2 + 2p - 3 > 0$.
Factoring the quadratic: $(p+3)(p-1) > 0$.
The inequality holds when $p < -3$ or $p > 1$.
Thus,the lines are distinct for all $p \in R - [-3, 1]$.

(C) Let the angle between the given pair of lines be $\theta$.
According to the given information,the area of the bigger sector is $5$ times the area of the smaller sector.
Let $A_1$ be the area of the smaller sector and $A_2$ be the area of the bigger sector.
$A_2 = 5 A_1 \Rightarrow \frac{1}{2}(\pi - \theta) r^2 = 5 \times \left(\frac{1}{2} \theta r^2\right)$
$\Rightarrow \pi - \theta = 5 \theta$ $\Rightarrow 6 \theta = \pi$ $\Rightarrow \theta = \frac{\pi}{6}$
The angle $\theta$ between the pair of lines $a x^2 + 2h x y + b y^2 = 0$ is given by $\tan \theta = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} \right|$.
Here,$a = l$,$h = l+m$,and $b = m$.
$\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} = \left| \frac{2 \sqrt{(l+m)^2 - lm}}{l+m} \right|$
Squaring both sides:
$\frac{1}{3} = \frac{4((l+m)^2 - lm)}{(l+m)^2}$
$(l+m)^2 = 12(l+m)^2 - 12 lm$
$11(l+m)^2 = 12 lm$
$\frac{lm}{(l+m)^2} = \frac{11}{12}$
Hence,option $(c)$ is correct.

(B) The given equation is $x^2 + 2hxy + 6y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$b = 6$,and the coefficient of $xy$ is $2h$.
Let the slopes of the lines be $m_1$ and $m_2$. Then $m_1 + m_2 = -\frac{2h}{6} = -\frac{h}{3}$ and $m_1 m_2 = \frac{1}{6}$.
Given that the angle $\theta$ between the lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,we have $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \frac{1}{7}$.
Using $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$,we get $(m_1 - m_2)^2 = \left(-\frac{h}{3}\right)^2 - 4\left(\frac{1}{6}\right) = \frac{h^2}{9} - \frac{2}{3} = \frac{h^2 - 6}{9}$.
Thus,$|m_1 - m_2| = \frac{\sqrt{h^2 - 6}}{3}$.
Substituting this into the tangent formula: $\frac{\frac{\sqrt{h^2 - 6}}{3}}{1 + \frac{1}{6}} = \frac{1}{7} \implies \frac{\sqrt{h^2 - 6}}{3} \times \frac{6}{7} = \frac{1}{7} \implies \frac{2\sqrt{h^2 - 6}}{7} = \frac{1}{7}$.
This gives $2\sqrt{h^2 - 6} = 1$,so $4(h^2 - 6) = 1$,which means $4h^2 - 24 = 1$,so $4h^2 = 25$,$h^2 = \frac{25}{4}$,$h = \pm \frac{5}{2}$.
Since the slopes are positive,$m_1 + m_2 = -\frac{h}{3} > 0$,so $h$ must be negative. Thus,$h = -\frac{5}{2}$.

(C) The given equation is $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$.
Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$.
Rearranging the terms: $x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha + y^2(\sin^2 \alpha - \sin^2 \alpha) = 0$.
This simplifies to: $-x^2 \cos 2\alpha + xy \sin 2\alpha = 0$.
For a pair of lines $Ax^2 + 2Hxy + By^2 = 0$ to be perpendicular,the condition is $A + B = 0$.
Here,$A = -\cos 2\alpha$ and $B = 0$.
Thus,$-\cos 2\alpha + 0 = 0$,which implies $\cos 2\alpha = 0$.
Since $\cos 2\alpha = 1 - 2\sin^2 \alpha = 0$,we get $\sin^2 \alpha = \frac{1}{2}$.
Then $\cos^2 \alpha = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{1/2}{1/2} = 1$.
Finally,$\sin^2 \alpha + \tan^2 \alpha = \frac{1}{2} + 1 = \frac{3}{2}$.

(A) The general equation of a pair of lines passing through the origin is given by $Ax^2 + 2Hxy + By^2 = 0$.
For these lines to be at right angles to each other,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$A + B = 0$.
In the given equation $3ax^2 + 5xy + (a^2 - 2)y^2 = 0$,we have $A = 3a$ and $B = a^2 - 2$.
Setting $A + B = 0$,we get $3a + a^2 - 2 = 0$,which is $a^2 + 3a - 2 = 0$.
Solving this quadratic equation for $a$ using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $a = \frac{-3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{-3 \pm \sqrt{17}}{2}$.
Since the discriminant $D = 17 > 0$,there are two distinct real values for $a$.
Thus,the number of values of $a$ is $2$.

(A) The equation of the pair of lines is $2x^2 + 3xy + ky^2 = 0$.
Dividing by $x^2$,we get $k(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$.
Let $m = \frac{y}{x}$ be the slope of the lines. Then $km^2 + 3m + 2 = 0$.
Given that one slope $m_1 = 2$,substituting this into the equation: $k(2)^2 + 3(2) + 2 = 0 \implies 4k + 6 + 2 = 0 \implies 4k = -8 \implies k = -2$.
The equation becomes $2x^2 + 3xy - 2y^2 = 0$.
Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 2$,$2h = 3 \implies h = \frac{3}{2}$,and $b = -2$.
The angle $\theta$ between the lines is given by $\tan \theta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$.
Here $a + b = 2 + (-2) = 0$.
Since the sum of coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular.
Therefore,$\theta = \frac{\pi}{2}$.

(B) Let the angle between the given pair of lines be $\theta$. The area of a sector with angle $\alpha$ is given by $\frac{1}{2}r^2\alpha$.
Given that the area of the bigger sector is $5$ times the area of the smaller sector:
$\frac{1}{2}r^2(\pi-\theta) = 5 \times \frac{1}{2}r^2\theta$
$\pi-\theta = 5\theta$
$6\theta = \pi \implies \theta = \frac{\pi}{6}$.
The angle $\theta$ between the lines represented by $ax^2+2hxy+by^2=0$ satisfies $\cos \theta = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$.
Substituting $\theta = \frac{\pi}{6}$:
$\cos \frac{\pi}{6} = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$
$\frac{\sqrt{3}}{2} = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$.

(B) The general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel lines if $h^2=ab$ and $bg^2=af^2$.
From the condition $h^2=ab$,we have $h=\sqrt{ab}$.
From the condition $bg^2=af^2$,we can write $\frac{g^2}{f^2}=\frac{a}{b}$.
Taking the square root on both sides,we get $\frac{g}{f}=\sqrt{\frac{a}{b}}$.
For parallel lines,the distance between them is given by $2\sqrt{\frac{g^2-ac}{a(a+b)}} = 2\sqrt{\frac{f^2-bc}{b(a+b)}}$.
This implies $\frac{g^2-ac}{a} = \frac{f^2-bc}{b}$.
Rearranging the terms,we get $\frac{g^2-ac}{f^2-bc} = \frac{a}{b}$.
Therefore,$\sqrt{\frac{g^2-ac}{f^2-bc}} = \sqrt{\frac{a}{b}}$.

(B) The equation of the pair of lines is $ax^2+4xy+2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=a$,$2H=4 \Rightarrow H=2$,and $B=2$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{H^2-AB}}{A+B} \right|$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1$.
$1 = \left| \frac{2\sqrt{2^2-a(2)}}{a+2} \right| = \left| \frac{2\sqrt{4-2a}}{a+2} \right|$.
Squaring both sides,we get $1 = \frac{4(4-2a)}{(a+2)^2}$.
$(a+2)^2 = 16-8a \Rightarrow a^2+4a+4 = 16-8a$.
$a^2+12a-12 = 0$.
Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $a = \frac{-12 \pm \sqrt{144 - 4(1)(-12)}}{2} = \frac{-12 \pm \sqrt{144+48}}{2} = \frac{-12 \pm \sqrt{192}}{2} = \frac{-12 \pm 8\sqrt{3}}{2} = -6 \pm 4\sqrt{3}$.

(D) The given equation is $3ax^2 - 16xy - (a^2 - 10)y^2 = 0$.
Comparing this with the general form $Ax^2 + 2Hxy + By^2 = 0$,we get $A = 3a$,$2H = -16$ (so $H = -8$),and $B = -(a^2 - 10)$.
For the equation to represent a pair of parallel lines,the condition is $H^2 = AB$.
Substituting the values: $(-8)^2 = 3a(-(a^2 - 10))$ $\Rightarrow 64 = -3a^3 + 30a$ $\Rightarrow 3a^3 - 30a + 64 = 0$.
For the equation to represent a pair of perpendicular lines,the condition is $A + B = 0$.
Substituting the values: $3a - (a^2 - 10) = 0$ $\Rightarrow 3a - a^2 + 10 = 0$ $\Rightarrow a^2 - 3a - 10 = 0$.
Factoring the quadratic: $(a - 5)(a + 2) = 0$,which gives $a = 5$ or $a = -2$.
Comparing these results with the given options,option $D$ is correct.

(C) Given the equation of the pair of lines is $H \equiv ax^2 - xy + by^2 = 0$.
Since the point $(1, -1)$ lies on $H = 0$,we substitute $x = 1$ and $y = -1$ into the equation:
$a(1)^2 - (1)(-1) + b(-1)^2 = 0$
$a + 1 + b = 0 \Rightarrow a + b = -1$
Comparing $ax^2 - xy + by^2 = 0$ with the general form $ax^2 + 2hxy + by^2 = 0$,we get $2h = -1$,so $h = -\frac{1}{2}$.
The formula for the acute angle $\theta$ between the pair of lines is $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Given $\tan \theta = 5$,we have:
$5 = \left| \frac{2\sqrt{(-\frac{1}{2})^2 - ab}}{-1} \right| = \left| \frac{2\sqrt{\frac{1}{4} - ab}}{-1} \right| = 2\sqrt{\frac{1}{4} - ab}$
Squaring both sides:
$25 = 4(\frac{1}{4} - ab)$ $\Rightarrow 25 = 1 - 4ab$ $\Rightarrow 4ab = -24$ $\Rightarrow ab = -6$.
Now,we need to find $a^2 + ab + b^2$.
$a^2 + ab + b^2 = (a + b)^2 - ab = (-1)^2 - (-6) = 1 + 6 = 7$.

(A) The equation of the pair of lines is $ax^2-4xy-2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=-4 \Rightarrow H=-2$,and $B=-2$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{H^2-AB}}{A+B} \right|$.
Substituting the values,$\tan \theta = \left| \frac{2\sqrt{(-2)^2 - a(-2)}}{a-2} \right| = \left| \frac{2\sqrt{4+2a}}{a-2} \right|$.
Given $\cos \theta = \frac{1}{5}$,we have $\sin \theta = \sqrt{1-\cos^2 \theta} = \sqrt{1-\frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{24}/5}{1/5} = \sqrt{24}$.
Equating the two expressions for $\tan \theta$: $\frac{2\sqrt{4+2a}}{|a-2|} = \sqrt{24}$.
Squaring both sides: $\frac{4(4+2a)}{(a-2)^2} = 24 \Rightarrow \frac{4+2a}{(a-2)^2} = 6$.
$4+2a = 6(a^2-4a+4) \Rightarrow 4+2a = 6a^2-24a+24$.
$6a^2-26a+20 = 0 \Rightarrow 3a^2-13a+10 = 0$.
Factoring the quadratic: $(3a-10)(a-1) = 0$.
So,$a_1=1$ and $a_2=\frac{10}{3}$.
Finally,$a_1+3a_2 = 1 + 3(\frac{10}{3}) = 1+10 = 11$.

(D) Given the equation of the pair of lines is $5x^2 + \frac{40}{3}xy + ky^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 5$,$2h = \frac{40}{3} \Rightarrow h = \frac{20}{3}$,and $b = k$.
Let the slopes of the lines be $m_1$ and $m_2$. We know that $m_1 + m_2 = -\frac{2h}{b} = -\frac{40}{3k}$ and $m_1m_2 = \frac{a}{b} = \frac{5}{k}$.
Given $m_1 = 3$,we have $3 + m_2 = -\frac{40}{3k}$ and $3m_2 = \frac{5}{k} \Rightarrow m_2 = \frac{5}{3k}$.
Substituting $m_2$ into the sum equation: $3 + \frac{5}{3k} = -\frac{40}{3k}$.
Multiplying by $3k$: $9k + 5 = -40$ $\Rightarrow 9k = -45$ $\Rightarrow k = -5$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Here,$a + b = 5 + (-5) = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular.
Therefore,$\theta = \frac{\pi}{2}$.

(C) The cosine of the angle $\theta$ between the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\cos \theta = \frac{|a+b|}{\sqrt{(a-b)^2 + 4h^2}}$.
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$: Here $a=5, h=\sqrt{7}, b=-1$. $\cos \theta = \frac{|5-1|}{\sqrt{(5-(-1))^2 + 4(\sqrt{7})^2}} = \frac{4}{\sqrt{36+28}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = \frac{1}{2}$. Thus,$A-III$.
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$: Here $a=1, h=\frac{\sqrt{11}}{2}, b=2$. $\cos \theta = \frac{|1+2|}{\sqrt{(1-2)^2 + 4(\frac{\sqrt{11}}{2})^2}} = \frac{3}{\sqrt{1+11}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. Thus,$B-I$.
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$: Here $a=1, h=\sqrt{2}, b=1$. $\cos \theta = \frac{|1+1|}{\sqrt{(1-1)^2 + 4(\sqrt{2})^2}} = \frac{2}{\sqrt{0+8}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$. Thus,$C-V$.
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$: Here $a=3, h=2\sqrt{2}, b=1$. $\cos \theta = \frac{|3+1|}{\sqrt{(3-1)^2 + 4(2\sqrt{2})^2}} = \frac{4}{\sqrt{4+32}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3}$. Thus,$D-IV$.
Therefore,the correct match is $A-III, B-I, C-V, D-IV$.

(D) The general equation of a pair of lines passing through the origin is $Ax^2 + 2Hxy + By^2 = 0$.
For the lines to be at right angles,the condition is $A + B = 0$.
Here,$A = \alpha^2 + 12|\alpha|$ and $B = 18 - 21|\alpha|$.
Setting $A + B = 0$,we get $\alpha^2 + 12|\alpha| + 18 - 21|\alpha| = 0$.
This simplifies to $\alpha^2 - 9|\alpha| + 18 = 0$.
Let $|\alpha| = t$,where $t \ge 0$. Then $t^2 - 9t + 18 = 0$.
Factoring the quadratic,we get $(t - 3)(t - 6) = 0$.
So,$t = 3$ or $t = 6$.
Since $|\alpha| = 3$,we have $\alpha = 3$ or $\alpha = -3$.
Since $|\alpha| = 6$,we have $\alpha = 6$ or $\alpha = -6$.
Thus,there are $4$ real values of $\alpha$.

(D) The angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Given $a = 12$,$b = 7$,and $\tan \theta = \frac{8}{19}$.
Substituting the values,we get $\frac{8}{19} = \left| \frac{2\sqrt{h^2 - 12 \times 7}}{12 + 7} \right|$.
$\frac{8}{19} = \frac{2\sqrt{h^2 - 84}}{19}$.
$8 = 2\sqrt{h^2 - 84}$.
$4 = \sqrt{h^2 - 84}$.
Squaring both sides,$16 = h^2 - 84$.
$h^2 = 100$.
$h = \pm 10$.

(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$.
This represents two lines: $x=1$ and $y=1$.
Since these lines are concurrent with $x+ay-3=0$,the point of intersection $(1,1)$ must satisfy the equation $x+ay-3=0$.
Substituting $(1,1)$ into $x+ay-3=0$,we get $1+a(1)-3=0$,which implies $a=2$.
Substituting $a=2$ into the second pair of lines,we get $2x^2-13xy-7y^2+x+23y-6=0$.
For a general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,the angle $\theta$ between the lines is given by $\tan \theta = \left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$.
Here $A=2, 2H=-13, B=-7$.
Thus,$\tan \theta = \left|\frac{2\sqrt{(-13/2)^2 - (2)(-7)}}{2-7}\right| = \left|\frac{2\sqrt{169/4 + 14}}{-5}\right| = \left|\frac{2\sqrt{225/4}}{-5}\right| = \left|\frac{2(15/2)}{-5}\right| = |-3| = 3$.
Since $\tan \theta = 3$,we have $\cos \theta = \frac{1}{\sqrt{1+3^2}} = \frac{1}{\sqrt{10}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$.

(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5}{2}, b=p, g=\frac{3}{2}, f=-4, c=2$.
The equation represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values: $1(p)(2) + 2(-4)(\frac{3}{2})(-\frac{5}{2}) - 1(-4)^2 - p(\frac{3}{2})^2 - 2(-\frac{5}{2})^2 = 0$.
$2p + 30 - 16 - \frac{9p}{4} - \frac{25}{2} = 0$.
Multiplying by $4$: $8p + 120 - 64 - 9p - 50 = 0$ $\Rightarrow -p + 6 = 0$ $\Rightarrow p=6$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{(-\frac{5}{2})^2 - 1(6)}}{1+6} \right| = \left| \frac{2\sqrt{\frac{25}{4}-6}}{7} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{7} \right| = \frac{2(\frac{1}{2})}{7} = \frac{1}{7}$.
Since $\tan \theta = \frac{1}{7}$,we have a right triangle with opposite side $1$ and adjacent side $7$. The hypotenuse is $\sqrt{1^2+7^2} = \sqrt{50}$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{50}}$.

(D) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$.
For this to represent a pair of parallel lines,it must satisfy $h^2=ab$ and $af^2=bg^2$.
Given $4x^2+12xy+9y^2+2gx+2fy-1=0$,we have $a=4, h=6, b=9$.
Check condition $h^2=ab$: $6^2 = 36$ and $4 \times 9 = 36$. This holds.
Now use $af^2=bg^2$: $4f^2=9g^2$,which implies $f^2/g^2 = 9/4$,so $f/g = \pm 3/2$.
Also,for parallel lines,the equation can be written as $(2x+3y+k_1)(2x+3y+k_2)=0$.
Expanding this: $4x^2+12xy+9y^2+2k_1x+2k_2x+3k_1y+3k_2y+k_1k_2=0$.
Comparing coefficients: $2g = 2(k_1+k_2) \implies g = k_1+k_2$ and $2f = 3(k_1+k_2) \implies 2f = 3g \implies f/g = 3/2$.
Also $k_1k_2 = -1$.
Since $g = k_1+k_2$ and $f = \frac{3}{2}g$,we have $f/g = 3/2$.
Substituting into the options,$\frac{f}{g} + \frac{g}{f} = \frac{3}{2} + \frac{2}{3} = \frac{9+4}{6} = \frac{13}{6}$.

(C) The general equation of a second-degree curve is given by $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.
Comparing this with the given equation $(a^2-3)x^2 + 16xy - 2ay^2 + 4x - 8y - 2 = 0$,we have:
$A = a^2-3$,$2H = 16 \implies H = 8$,$B = -2a$.
For the equation to represent a pair of perpendicular lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$A + B = 0$.
Substituting the values,we get $(a^2-3) + (-2a) = 0$.
$a^2 - 2a - 3 = 0$.
Factoring the quadratic equation: $(a-3)(a+1) = 0$.
Thus,$a = 3$ or $a = -1$.
Checking the condition for a pair of lines (the determinant $\Delta = 0$):
$\Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$.
For $a = 3$: $A = 6, B = -6, H = 8, G = 2, F = -4, C = -2$.
$\Delta = (6)(-6)(-2) + 2(-4)(2)(8) - (6)(-4)^2 - (-6)(2)^2 - (-2)(8)^2 = 72 - 128 - 96 + 24 + 128 = 0$.
For $a = -1$: $A = -2, B = 2, H = 8, G = 2, F = -4, C = -2$.
$\Delta = (-2)(2)(-2) + 2(-4)(2)(8) - (-2)(-4)^2 - (2)(2)^2 - (-2)(8)^2 = 8 - 128 + 32 - 8 + 128 = 32 \neq 0$.
Since $a = -1$ does not satisfy the condition for a pair of lines,the only valid value is $a = 3$.

(A) The given equation of the pair of straight lines is $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$.
Rearranging the terms,we have $-3x^2 - 8xy + 3y^2 - 29x + 3y - 18 = 0$.
Comparing this with the general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = -3$ and $b = 3$.
Since the sum of the coefficients of $x^2$ and $y^2$ is $a + b = -3 + 3 = 0$,the lines are perpendicular to each other.
Therefore,the angle between the pair of straight lines is $90^{\circ}$.

(D) The equation of the pair of lines is $ax^2 + 4xy - 2y^2 = 0$. Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we get $A = a$,$2H = 4 \implies H = 2$,and $B = -2$.
Let $\theta$ be the angle between the lines. Then $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A + B} \right|$.
Given $\tan \theta = 2\sqrt{10}$,so $2\sqrt{10} = \left| \frac{2\sqrt{2^2 - a(-2)}}{a - 2} \right| = \left| \frac{2\sqrt{4 + 2a}}{a - 2} \right|$.
Squaring both sides: $40 = \frac{4(4 + 2a)}{(a - 2)^2} \implies 10 = \frac{4 + 2a}{a^2 - 4a + 4}$.
$10a^2 - 40a + 40 = 4 + 2a \implies 10a^2 - 42a + 36 = 0 \implies 5a^2 - 21a + 18 = 0$.
Solving for $a$: $5a^2 - 15a - 6a + 18 = 0 \implies 5a(a - 3) - 6(a - 3) = 0 \implies (5a - 6)(a - 3) = 0$.
Since $a \in \mathbb{Z}$,we have $a = 3$.
The equation of the lines is $3x^2 + 4xy - 2y^2 = 0$. Dividing by $x^2$ and letting $m = y/x$,we get $-2m^2 + 4m + 3 = 0$,or $2m^2 - 4m - 3 = 0$.
The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$,which is $-3/2$.