General Second-Degree equation Questions in English

(C) To check for symmetry about the line $y = x$,we interchange $x$ and $y$ in the given equation $x^3 + y^3 = 3axy$.
Replacing $x$ with $y$ and $y$ with $x$,we get $y^3 + x^3 = 3ayx$,which is equivalent to $x^3 + y^3 = 3axy$.
Since the equation remains unchanged,the curve is symmetric about the line $y = x$.

(C) Since the point $(2, -3)$ lies on the curve $kx^2 - 3y^2 + 2x + y - 2 = 0$,it must satisfy the equation.
Substituting $x = 2$ and $y = -3$ into the equation:
$k(2)^2 - 3(-3)^2 + 2(2) + (-3) - 2 = 0$
$4k - 3(9) + 4 - 3 - 2 = 0$
$4k - 27 + 4 - 5 = 0$
$4k - 28 = 0$
$4k = 28$
$k = 7$

(D) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing with $2x^2 - 72xy + 23y^2 - 4x - 28y - 48 = 0$,we have $a=2, h=-36, b=23, g=-2, f=-14, c=-48$.
The centre $(x, y)$ is given by the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
$4x - 72y - 4 = 0 \implies x - 18y = 1$ $(i)$
$-72x + 46y - 28 = 0 \implies -36x + 23y = 14$ (ii)
Multiplying $(i)$ by $36$: $36x - 648y = 36$.
Adding to (ii): $(23 - 648)y = 14 + 36 \implies -625y = 50 \implies y = -\frac{50}{625} = -\frac{2}{25}$.
Substituting $y$ in $(i)$: $x - 18(-\frac{2}{25}) = 1 \implies x + \frac{36}{25} = 1 \implies x = 1 - \frac{36}{25} = -\frac{11}{25}$.
Thus,the centre is $\left( -\frac{11}{25}, -\frac{2}{25} \right)$.

(A) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$,we get:
$a = 14, h = -2, b = 11, g = -22, f = -29, c = 71$.
The centre $(x, y)$ of the conic is given by the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
$\frac{\partial}{\partial x} = 28x - 4y - 44 = 0 \implies 7x - y = 11$ $(1)$
$\frac{\partial}{\partial y} = -4x + 22y - 58 = 0 \implies -2x + 11y = 29$ $(2)$
Multiplying $(1)$ by $11$,we get $77x - 11y = 121$.
Adding this to $(2)$,we get $75x = 150$,so $x = 2$.
Substituting $x = 2$ in $(1)$,$7(2) - y = 11 \implies 14 - y = 11 \implies y = 3$.
Thus,the centre is $(2, 3)$.

(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation $x^2 - 2xy + y^2 + 3x + 2 = 0$,we have $a = 1$,$h = -1$,$b = 1$,$g = 3/2$,$f = 0$,and $c = 2$.
First,we check the condition $h^2 - ab$:
$h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$.
Since $h^2 - ab = 0$,the conic section represents a parabola (provided it is not a pair of parallel lines).
To confirm it is not a degenerate conic,we check the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$:
$\Delta = (1)(1)(2) + 2(0)(3/2)(-1) - (1)(0)^2 - (1)(3/2)^2 - (2)(-1)^2$
$\Delta = 2 + 0 - 0 - 9/4 - 2 = -9/4 \neq 0$.
Since $\Delta \neq 0$ and $h^2 - ab = 0$,the equation represents a parabola.

(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation $16x^2 + 8xy + y^2 - 74x - 78y + 212 = 0$,we get $a = 16$,$h = 4$,and $b = 1$.
We check the condition $h^2 - ab = (4)^2 - (16)(1) = 16 - 16 = 0$.
Since $h^2 - ab = 0$,the curve represents a parabola.

(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation,we have $a = 14$,$h = -2$,$b = 11$,$g = -22$,$f = -29$,and $c = 71$.
First,we calculate the discriminant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$:
$\Delta = (14)(11)(71) + 2(-29)(-22)(-2) - 14(-29)^2 - 11(-22)^2 - 71(-2)^2$.
$\Delta = 10934 - 2552 - 11774 - 5324 - 284 = -9000 \neq 0$.
Since $\Delta \neq 0$,the equation represents a conic section.
Next,we check the condition $h^2 - ab$:
$h^2 - ab = (-2)^2 - (14)(11) = 4 - 154 = -150$.
Since $h^2 - ab < 0$,the conic section is an ellipse.

(D) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
For a hyperbola,the discriminant $\Delta \neq 0$ and $h^2 > ab$.
For option $(d)$,the equation is $x^2 - y^2 = 0$,which can be written as $(x - y)(x + y) = 0$.
This represents a pair of intersecting straight lines,not a hyperbola,because $\Delta = 0$ in this case.

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation $x^2 + 4xy + y^2 + 2x + 4y + 2 = 0$,we have $a = 1, h = 2, b = 1, g = 1, f = 2, c = 2$.
First,we calculate the discriminant $h^2 - ab = (2)^2 - (1)(1) = 4 - 1 = 3$.
Since $h^2 - ab > 0$,the conic section is a hyperbola.
Next,we check the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = (1)(1)(2) + 2(2)(1)(1) - (1)(2)^2 - (1)(1)^2 - (2)(2)^2 = 2 + 4 - 4 - 1 - 8 = -7$.
Since $\Delta \neq 0$ and $h^2 - ab > 0$,the equation represents a hyperbola.

(A) Let $f(x, y) = 14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.
Taking partial derivatives with respect to $x$ and $y$:
$\frac{\partial f}{\partial x} = 28x - 4y - 44$ and $\frac{\partial f}{\partial y} = -4x + 22y - 58$.
For the center,set $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$:
$28x - 4y - 44 = 0 \implies 7x - y = 11$ (Equation $I$)
$-4x + 22y - 58 = 0 \implies -2x + 11y = 29$ (Equation $II$)
Solving the system of linear equations:
From $I$,$y = 7x - 11$.
Substitute into $II$: $-2x + 11(7x - 11) = 29 \implies -2x + 77x - 121 = 29 \implies 75x = 150 \implies x = 2$.
Then $y = 7(2) - 11 = 3$.
Therefore,the center is $(2, 3)$.

(C) Comparing the given equation with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get:
$a = 13, h = -9, b = 37, g = 1, f = 7, c = -2$.
First,we calculate the discriminant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$:
$\Delta = (13)(37)(-2) + 2(7)(1)(-9) - 13(7)^2 - 37(1)^2 - (-2)(-9)^2$
$\Delta = -962 - 126 - 637 - 37 + 162 = -1600 \neq 0$.
Since $\Delta \neq 0$,the conic is non-degenerate.
Next,we check the condition $h^2 - ab$:
$h^2 = (-9)^2 = 81$ and $ab = 13 \times 37 = 481$.
$h^2 - ab = 81 - 481 = -400 < 0$,which implies $ab - h^2 > 0$.
Since $\Delta \neq 0$ and $h^2 - ab < 0$,the equation represents an ellipse.

(B) The general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a conic section.
Here,$\Delta$ is the determinant of the matrix $\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$.
For the equation to represent an ellipse,the following conditions must be satisfied:
$1$. $\Delta \neq 0$ (non-degenerate conic).
$2$. $h^2 - ab < 0$,which implies $h^2 < ab$.

(D) Given the equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$ with $k > 1$.
Since $k > 1$,we have $1-k < 0$ and $1+k > 2$.
Let $a^2 = k-1$ and $b^2 = k+1$.
Then the equation becomes $\frac{x^2}{-(k-1)} - \frac{y^2}{k+1} = 1$,which is $-\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Multiplying by $-1$,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = -1$.
Since the sum of two squares cannot be negative,this equation represents an empty set (no real locus).
However,if the question implies the standard form of a hyperbola where the signs of the denominators differ,we check the discriminant $B^2 - 4AC$.
Here $A = \frac{1}{1-k}$ and $C = -\frac{1}{1+k}$.
$B^2 - 4AC = 0 - 4(\frac{1}{1-k})(-\frac{1}{1+k}) = \frac{4}{1-k^2}$.
Since $k > 1$,$k^2 > 1$,so $1-k^2 < 0$.
Thus,$B^2 - 4AC < 0$,which typically represents an ellipse or an empty set.
Given the structure,it does not represent a hyperbola for $k > 1$.

(B) The given equation is $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 5$,$2h = 8$ (so $h = 4$),and $b = 5$.
The angle of rotation $\theta$ required to eliminate the $xy$ term is given by the formula $\tan(2\theta) = \frac{2h}{a - b}$.
Substituting the values,we have $\tan(2\theta) = \frac{8}{5 - 5} = \frac{8}{0}$,which is undefined.
Therefore,$2\theta = \frac{\pi}{2}$,which implies $\theta = \frac{\pi}{4}$.

(C) general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ represents a parabola if the discriminant $B^2 - 4AC = 0$.
For option $A$: $A=4, B=-12, C=9$. Discriminant $B^2 - 4AC = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. This is a parabola.
For option $B$: $A=4, B=-12, C=9$. Discriminant $B^2 - 4AC = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. This is a parabola.
For option $C$: $A=2, B=-4, C=1$. Discriminant $B^2 - 4AC = (-4)^2 - 4(2)(1) = 16 - 8 = 8$. Since $8 \neq 0$,this does not represent a parabola (it represents a hyperbola as $B^2 - 4AC > 0$).

(D) Given the equation: $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ where $r > 1$.
Since $r > 1$,let $r - 1 = p$,where $p > 0$. Then $1 - r = -p$.
Substituting this into the equation,we get: $\frac{x^2}{-p} - \frac{y^2}{1 + r} = 1$.
Multiplying by $-1$,we get: $\frac{x^2}{p} + \frac{y^2}{1 + r} = -1$.
Since the sum of two squares divided by positive constants equals a negative value $(-1)$,there are no real values of $(x, y)$ that satisfy this equation.
Therefore,the equation represents an imaginary ellipse.

(C) Given the curve $3x^2 + 4xy + 5y^2 - 4 = 0$. Differentiating with respect to $x$,we get $6x + 4y + 4x\frac{dy}{dx} + 10y\frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{6x + 4y}{4x + 10y} = -\frac{3x + 2y}{2x + 5y}$.
For point $P$ on the line $y = -\frac{3}{2}x$,we have $2y = -3x$,so $3x + 2y = 0$. Substituting this into the derivative,$\left. \frac{dy}{dx} \right|_P = 0$.
For point $Q$ on the line $y = -\frac{2}{5}x$,we have $5y = -2x$,so $2x + 5y = 0$. Substituting this into the derivative,$\left. \frac{dy}{dx} \right|_Q = \infty$.
Since the slope at $P$ is $0$ (horizontal tangent) and the slope at $Q$ is $\infty$ (vertical tangent),the tangents are perpendicular to each other.

(C) For the curves $y = ax^2 + bx + c$ and $y = px^2 + qx + r$ to not intersect,the equation $ax^2 + bx + c = px^2 + qx + r$ must have no real roots.
This implies $(a - p)x^2 + (b - q)x + (c - r) = 0$ has no real roots,so the discriminant $D < 0$.
$D = (b - q)^2 - 4(a - p)(c - r) < 0$.
Given $a, b, c, p, q, r \in \{1, 2, \dots, 10\}$,we want to maximize $(aq - bp)^2 + (c - r)^2$.
If $aq = bp$,then $(aq - bp)^2 = 0$,and we maximize $(c - r)^2$. With $c, r \in \{1, 10\}$,max $(c - r)^2 = (10 - 1)^2 = 81$.
However,if $aq \neq bp$,we can achieve larger values. For $a=1, p=10, b=10, q=1$,$aq - bp = 1 - 100 = -99$,which is too large.
Considering the constraint $D < 0$,for $a=1, p=2, b=1, q=2$,$D = (1-2)^2 - 4(1-2)(c-r) = 1 + 4(c-r) < 0$,so $c-r < -1/4$. If $c=1, r=10$,$c-r = -9$,$D = 1 + 4(-9) = -35 < 0$.
Then $(aq - bp)^2 + (c - r)^2 = (1(2) - 10(1))^2 + (1 - 10)^2 = (-8)^2 + (-9)^2 = 64 + 81 = 145$.
Testing $a=1, p=10, b=1, q=10$,$aq-bp = 10-10=0$. Testing $a=1, p=2, b=10, q=1$,$aq-bp = 1-20 = -19$,$(aq-bp)^2 = 361$ (too large).
The maximum value under the condition $D < 0$ is $162$.

(B) The general second-degree equation is given by $Ax^2 + 2hxy + By^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$,we have $A = 5$,$h = 4$,and $B = 5$.
To eliminate the $xy$ term,the axes must be rotated by an angle $\theta$ such that $\tan(2\theta) = \frac{2h}{A - B}$.
Substituting the values,we get $\tan(2\theta) = \frac{2(4)}{5 - 5} = \frac{8}{0} = \infty$.
Since $\tan(2\theta) = \infty$,we have $2\theta = \frac{\pi}{2}$.
Therefore,$\theta = \frac{\pi}{4}$.

(A) Let $P(x, y)$. The distances from $L_1$ and $L_2$ are $d_1 = \frac{|x\sqrt{2} + y - 1|}{\sqrt{2+1}} = \frac{|x\sqrt{2} + y - 1|}{\sqrt{3}}$ and $d_2 = \frac{|x\sqrt{2} - y + 1|}{\sqrt{2+1}} = \frac{|x\sqrt{2} - y + 1|}{\sqrt{3}}$.
Given $d_1 d_2 = \lambda^2$,so $\frac{|(x\sqrt{2})^2 - (y-1)^2|}{3} = \lambda^2$,which gives $|2x^2 - (y-1)^2| = 3\lambda^2$.
For the line $y = 2x + 1$,substitute $y-1 = 2x$ into the locus equation: $|2x^2 - (2x)^2| = 3\lambda^2 \Rightarrow |-2x^2| = 3\lambda^2 \Rightarrow x^2 = \frac{3\lambda^2}{2}$.
The points $R$ and $S$ have $x$-coordinates $x_1 = \sqrt{\frac{3\lambda^2}{2}}$ and $x_2 = -\sqrt{\frac{3\lambda^2}{2}}$.
The distance $RS = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = \sqrt{(x_1-x_2)^2 + (2(x_1-x_2))^2} = \sqrt{5}|x_1-x_2| = \sqrt{5} \cdot 2\sqrt{\frac{3\lambda^2}{2}} = \sqrt{5} \cdot \sqrt{6\lambda^2} = \sqrt{30\lambda^2}$.
Given $\sqrt{30\lambda^2} = \sqrt{270} \Rightarrow 30\lambda^2 = 270 \Rightarrow \lambda^2 = 9$.
The midpoint $T$ of $RS$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (0, 1)$.
The slope of $RS$ is $2$,so the slope of the perpendicular bisector is $-\frac{1}{2}$.
The equation of the perpendicular bisector is $y - 1 = -\frac{1}{2}(x - 0) \Rightarrow x + 2y = 2 \Rightarrow x = 2 - 2y$.
Substitute $x = 2(1-y)$ into $|2x^2 - (y-1)^2| = 3\lambda^2 = 27$:
$|2(4(1-y)^2) - (y-1)^2| = 27 \Rightarrow |8(y-1)^2 - (y-1)^2| = 27 \Rightarrow 7(y-1)^2 = 27 \Rightarrow (y-1)^2 = \frac{27}{7}$.
The distance $D = (x_1^{\prime}-x_2^{\prime})^2 + (y_1^{\prime}-y_2^{\prime})^2 = 5(y_1^{\prime}-y_2^{\prime})^2 = 5(4(y-1)^2) = 20 \cdot \frac{27}{7} = \frac{540}{7} \approx 77.14$.

(B) Let the given equation be $f(x, y) = 2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$. \\ To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y = Y + k$. \\ The equation becomes $2(X+h)^2 + 4(X+h)(Y+k) + 5(Y+k)^2 - 4(X+h) - 22(Y+k) + 29 = 0$. \\ Expanding this,the linear terms in $X$ and $Y$ must vanish for the equation to be homogeneous. \\ The partial derivatives with respect to $x$ and $y$ are: \\ $f_x = 4x + 4y - 4 = 0 \implies x + y = 1$ \\ $f_y = 4x + 10y - 22 = 0 \implies 2x + 5y = 11$ \\ Solving these equations: \\ From the first,$x = 1 - y$. Substituting into the second: $2(1 - y) + 5y = 11 \implies 2 - 2y + 5y = 11 \implies 3y = 9 \implies y = 3$. \\ Then $x = 1 - 3 = -2$. \\ Thus,the origin must be shifted to $(-2, 3)$.

(A) 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 lines,the determinant must be zero:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ with the general form,we have $a=2, h=2, b=-p, g=2, f=q/2, c=1$.
Substituting these into the determinant condition:
$2(-p)(1) + 2(q/2)(2)(2) - 2(q/2)^2 - (-p)(2)^2 - 1(2)^2 = 0$.
$-2p + 4q - q^2/2 + 4p - 4 = 0$.
Multiplying by $-2$: $q^2 - 8q - 4p + 8 = 0$.
For $q$ to be an integer,the discriminant of this quadratic in $q$ must be a perfect square:
$D = (-8)^2 - 4(1)(8 - 4p) = 64 - 32 + 16p = 32 + 16p = 16(2 + p)$.
For $D$ to be a perfect square,$2 + p$ must be a perfect square $k^2$.
Given $p \in [-5, 5]$,$2 + p$ can take values from $-3$ to $7$.
The perfect squares in this range are $0, 1, 4$.
$2 + p = 0 \Rightarrow p = -2$.
$2 + p = 1 \Rightarrow p = -1$.
$2 + p = 4 \Rightarrow p = 2$.
Thus,the possible integral values of $p$ are $\{-2, -1, 2\}$.
The number of such values is $3$.

(A) Given equations are $x=t^2+t+1$ and $y=t^2-t+1$.
Adding the two equations: $x+y = 2t^2+2 = 2(t^2+1)$.
Subtracting the two equations: $x-y = 2t$,which implies $t = \frac{x-y}{2}$.
Substitute $t$ into the expression for $x+y$:
$x+y = 2\left(\left(\frac{x-y}{2}\right)^2 + 1\right)$
$x+y = 2\left(\frac{x^2+y^2-2xy}{4} + 1\right)$
$x+y = \frac{x^2+y^2-2xy}{2} + 2$
Multiplying by $2$: $2x+2y = x^2+y^2-2xy+4$
Rearranging the terms: $x^2-2xy+y^2-2x-2y+4=0$.

(C) The given equation is $2x^2+4xy-6y^2+2x+8y+1=0$.
Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=-6, g=1, f=4, c=1$.
To remove the first-degree terms,the origin must be shifted to the point $(h_0, k_0)$ given by the intersection of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
These are $4x+4y+2=0$ and $4x-12y+8=0$.
Solving these equations:
$2x+2y=-1$
$2x-6y=-4$
Subtracting the two equations: $8y=3 \implies y=\frac{3}{8}$.
Substituting $y=\frac{3}{8}$ into $2x+2y=-1$:
$2x+2(\frac{3}{8})=-1 \implies 2x+\frac{3}{4}=-1 \implies 2x=-\frac{7}{4} \implies x=-\frac{7}{8}$.
Thus,the required point is $\left(-\frac{7}{8}, \frac{3}{8}\right)$.

(D) Let $PQ$ subtend the angle $\theta$ at the origin,so $\angle POQ = \theta$.
Let the slope of the line be $m$.
The equation of the line passing through $(1, -2)$ is $y + 2 = m(x - 1)$.
Rearranging,we get $mx - y = m + 2$,or $\frac{mx - y}{m + 2} = 1$ ...$(i)$
The given curve is $3x^2 - y^2 - 2x + 4y = 0$ ...(ii)
To find the angle subtended at the origin,we homogenize the equation of the curve using the line equation:
$3x^2 - y^2 - 2x(1) + 4y(1) = 0$
Substituting $1 = \frac{mx - y}{m + 2}$:
$3x^2 - y^2 - 2x\left(\frac{mx - y}{m + 2}\right) + 4y\left(\frac{mx - y}{m + 2}\right) = 0$
Multiplying by $(m + 2)$:
$3(m + 2)x^2 - (m + 2)y^2 - 2x(mx - y) + 4y(mx - y) = 0$
$(3m + 6)x^2 - (m + 2)y^2 - 2mx^2 + 2xy + 4mxy - 4y^2 = 0$
$(m + 6)x^2 + (4m + 2)xy - (m + 6)y^2 = 0$
For the angle subtended to be $90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
Coefficient of $x^2$ + Coefficient of $y^2 = (m + 6) + (-(m + 6)) = 0$.
Since the sum is zero,the angle subtended by $PQ$ at the origin is $90^{\circ}$.

(D) The given curve is $3x^2-y^2-2x+4y=0$.
Let the chord passing through $(1,-2)$ have slope $m$. The equation of the line is $y+2=m(x-1)$,which can be written as $\frac{mx-y}{m+2}=1$.
To find the joint equation of the lines $OP$ and $OQ$,we homogenize the curve equation using the line equation:
$3x^2-y^2-(2x-4y)\left(\frac{mx-y}{m+2}\right)=0$.
Multiplying by $(m+2)$,we get:
$(m+2)(3x^2-y^2)-(2x-4y)(mx-y)=0$.
Expanding this:
$3mx^2+6x^2-my^2-2y^2-(2mx^2-2xy-4mxy+4y^2)=0$.
Grouping terms:
$x^2(3m+6-2m) + xy(2+4m) + y^2(-m-2-4) = 0$.
$x^2(m+6) + xy(4m+2) + y^2(-m-6) = 0$.
The equation is of the form $ax^2+2hxy+by^2=0$.
Here,$a = m+6$ and $b = -(m+6)$.
Since $a+b = (m+6) - (m+6) = 0$,the lines $OP$ and $OQ$ are perpendicular.
Therefore,$\theta = 90^{\circ}$.

(A) Let $f(x) = x^2 - 5ax + 6a$. For both roots to be greater than $1$,the following conditions must be satisfied:
$1$. Discriminant $D \ge 0$:
$D = (-5a)^2 - 4(1)(6a) = 25a^2 - 24a \ge 0 \implies a(25a - 24) \ge 0$.
This gives $a \in (- \infty, 0] \cup [\frac{24}{25}, \infty)$.
$2$. Vertex $x_v > 1$:
$x_v = -\frac{b}{2a} = \frac{5a}{2} > 1 \implies a > \frac{2}{5}$.
$3$. $f(1) > 0$:
$f(1) = 1^2 - 5a(1) + 6a = 1 + a > 0 \implies a > -1$.
Taking the intersection of all three conditions:
$a \in (- \infty, 0] \cup [\frac{24}{25}, \infty)$ $AND$ $a > \frac{2}{5}$ $AND$ $a > -1$.
The intersection is $a \in [\frac{24}{25}, \infty)$.

(A) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. $\ldots(i)$
Rotating the axes by $\theta = \frac{\pi}{4}$ implies $x = X\cos\theta - Y\sin\theta = \frac{X-Y}{\sqrt{2}}$ and $y = X\sin\theta + Y\cos\theta = \frac{X+Y}{\sqrt{2}}$.
Substituting these into $(i)$ gives the transformed equation:
$\frac{a}{2}(X-Y)^2 + h(X^2-Y^2) + \frac{b}{2}(X+Y)^2 + \sqrt{2}g(X-Y) + \sqrt{2}f(X+Y) + c = 0$.
Expanding and grouping terms:
$(\frac{a+b}{2}+h)X^2 + (b-a)XY + (\frac{a+b}{2}-h)Y^2 + \sqrt{2}(g+f)X + \sqrt{2}(f-g)Y + c = 0$.
Comparing with $Y^2+XY-X=0$,we get:
$1) \frac{a+b}{2}+h = 0$
$2) b-a = 1$
$3) \frac{a+b}{2}-h = 1$
$4) \sqrt{2}(g+f) = -1$
$5) f-g = 0$
From $(1)$ and $(3)$,$a+b=1$ and $h=-\frac{1}{2}$.
From $(2)$,$b-a=1$. Solving $a+b=1$ and $b-a=1$ gives $b=1, a=0$.
From $(4)$ and $(5)$,$f=g=-\frac{1}{2\sqrt{2}}$.
Now,$(h^2-ab)-2gf = ((-\frac{1}{2})^2 - (0)(1)) - 2(-\frac{1}{2\sqrt{2}})(-\frac{1}{2\sqrt{2}}) = \frac{1}{4} - 2(\frac{1}{8}) = \frac{1}{4} - \frac{1}{4} = 0$.

(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$.
Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=5, g=-2, f=-11, c=7$.
To eliminate the first-degree terms,the origin $(0,0)$ must be shifted to the point $(h', k')$ given by the intersection of the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
$\frac{\partial}{\partial x} = 4x+4y-4 = 0 \implies x+y=1$
$\frac{\partial}{\partial y} = 4x+10y-22 = 0 \implies 2x+5y=11$
Solving these equations:
From the first equation,$x = 1-y$.
Substituting into the second: $2(1-y)+5y=11 \implies 2-2y+5y=11 \implies 3y=9 \implies y=3$.
Then $x = 1-3 = -2$.
Thus,the origin should be shifted to $(-2, 3)$.

(B) The given equation is $16 x^2+y^2+8 x y-74 x-78 y+212=0$.
Comparing this with the general second-degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$,we get:
$a=16, b=1, h=4, g=-37, f=-39, c=212$.
Now,we calculate the discriminant $D = a b-h^2$:
$D = (16)(1)-(4)^2 = 16-16 = 0$.
Since $a b-h^2=0$,the given equation represents a parabola.

(A) The general equation of a second-degree curve is given by $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$.
Comparing the given equation $8 x^2+12 y^2-4 x+4 y-1=0$ with the general form,we have $a=8$,$b=12$,and $h=0$.
The nature of the conic section is determined by the discriminant $h^2-a b$.
Here,$h^2-a b = (0)^2 - (8)(12) = -96$.
Since $h^2-a b < 0$ and $a \neq b$,the equation represents an ellipse.

(B) Given curve is $x^{2}+4xy+8y^{2}=64$ ... $(i)$
Differentiating with respect to $x$,we get:
$2x + 4(y + x\frac{dy}{dx}) + 16y\frac{dy}{dx} = 0$
$2x + 4y + (4x + 16y)\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{x+2y}{2(x+4y)}$
Since the tangents are parallel to the $x$-axis,the slope $\frac{dy}{dx} = 0$.
This implies $x + 2y = 0$,so $x = -2y$ ... (ii)
Substituting $x = -2y$ into the original equation $(i)$:
$(-2y)^{2} + 4(-2y)y + 8y^{2} = 64$
$4y^{2} - 8y^{2} + 8y^{2} = 64$
$4y^{2} = 64$
$y^{2} = 16 \Rightarrow y = \pm 4$
If $y = 4$,then $x = -2(4) = -8$.
If $y = -4$,then $x = -2(-4) = 8$.
Thus,the required points are $(-8, 4)$ and $(8, -4)$.