Transformation of axes Questions in English

(C) Let $\theta = \tan^{-1}\left(\frac{3}{4}\right)$ be the angle of rotation. Then $\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$.
When the axes are rotated by an angle $\theta$,the transformation equations are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
Substituting the values of $\sin \theta$ and $\cos \theta$,we get $x = \frac{4X - 3Y}{5}$ and $y = \frac{3X + 4Y}{5}$.
Substituting these into the original equation $x^2 + y^2 = 9$:
$\left(\frac{4X - 3Y}{5}\right)^2 + \left(\frac{3X + 4Y}{5}\right)^2 = 9$
$\frac{16X^2 + 9Y^2 - 24XY + 9X^2 + 16Y^2 + 24XY}{25} = 9$
$\frac{25X^2 + 25Y^2}{25} = 9$
$X^2 + Y^2 = 9$.
Thus,the equation remains unchanged.

(C) The general equation of the second degree is $ax^2 + 2hxy + by^2 = 0$.
Comparing this with $x^2 + 2xy - y^2 = 0$,we get $a = 1$,$h = 1$,and $b = -1$.
The angle of rotation $\theta$ required to remove the $xy$ term is given by $\tan(2\theta) = \frac{2h}{a-b}$.
Substituting the values,$\tan(2\theta) = \frac{2(1)}{1 - (-1)} = \frac{2}{2} = 1$.
Since $\tan(2\theta) = 1$,we have $2\theta = \frac{\pi}{4}$.
Therefore,$\theta = \frac{\pi}{8}$.

(A) Let the original coordinates be $(X, Y)$ and the new coordinates be $(x, y)$.
Given the rotation angle $\theta = \frac{\pi}{4}$,the transformation equations are:
$X = x \cos\theta - y \sin\theta = x \frac{1}{\sqrt{2}} - y \frac{1}{\sqrt{2}} = \frac{x-y}{\sqrt{2}}$
$Y = x \sin\theta + y \cos\theta = x \frac{1}{\sqrt{2}} + y \frac{1}{\sqrt{2}} = \frac{x+y}{\sqrt{2}}$
Substituting these into the transformed equation $x^2+y^2-6x+8y+21=0$ is incorrect because the question asks for the original equation given the transformed one.
Actually,if the original equation is $f(X, Y) = 0$,the transformed equation is $f(x \cos\theta - y \sin\theta, x \sin\theta + y \cos\theta) = 0$.
Given $x^2+y^2-6x+8y+21=0$ is the transformed equation,we substitute $x = X \cos\theta + Y \sin\theta$ and $y = -X \sin\theta + Y \cos\theta$ where $\theta = \frac{\pi}{4}$.
$x = \frac{X+Y}{\sqrt{2}}, y = \frac{-X+Y}{\sqrt{2}}$
Substituting into $x^2+y^2-6x+8y+21=0$:
$(\frac{X+Y}{\sqrt{2}})^2 + (\frac{-X+Y}{\sqrt{2}})^2 - 6(\frac{X+Y}{\sqrt{2}}) + 8(\frac{-X+Y}{\sqrt{2}}) + 21 = 0$
$\frac{X^2+Y^2+2XY}{2} + \frac{X^2+Y^2-2XY}{2} - \frac{6X+6Y}{\sqrt{2}} + \frac{-8X+8Y}{\sqrt{2}} + 21 = 0$
$X^2+Y^2 - \frac{14X}{\sqrt{2}} + \frac{2Y}{\sqrt{2}} + 21 = 0$
$X^2+Y^2 - 7\sqrt{2}X + \sqrt{2}Y + 21 = 0$.

(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$. The rotation transformation is given by:
$x = X \cos \frac{\pi}{6} - Y \sin \frac{\pi}{6} = \frac{\sqrt{3}X - Y}{2}$
$y = X \sin \frac{\pi}{6} + Y \cos \frac{\pi}{6} = \frac{X + \sqrt{3}Y}{2}$
Substituting these into the equation $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$:
$\sqrt{3} \left( \frac{\sqrt{3}X - Y}{2} \right)^2 - 4 \left( \frac{\sqrt{3}X - Y}{2} \right) \left( \frac{X + \sqrt{3}Y}{2} \right) + \sqrt{3} \left( \frac{X + \sqrt{3}Y}{2} \right)^2 = 0$
$\frac{\sqrt{3}}{4} (3X^2 - 2\sqrt{3}XY + Y^2) - \frac{4}{4} (\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) + \frac{\sqrt{3}}{4} (X^2 + 2\sqrt{3}XY + 3Y^2) = 0$
Multiplying by $4$:
$3\sqrt{3}X^2 - 6XY + \sqrt{3}Y^2 - 4\sqrt{3}X^2 - 8XY + 4\sqrt{3}Y^2 + \sqrt{3}X^2 + 6XY + 3\sqrt{3}Y^2 = 0$
Combining like terms:
$(3\sqrt{3} - 4\sqrt{3} + \sqrt{3})X^2 + (-6 - 8 + 6)XY + (\sqrt{3} + 4\sqrt{3} + 3\sqrt{3})Y^2 = 0$
$0X^2 - 8XY + 8\sqrt{3}Y^2 = 0$
Dividing by $-8$:
$XY - \sqrt{3}Y^2 = 0$
Thus,the transformed equation is $\sqrt{3}Y^2 - XY = 0$.

(C) Step $1$: Under the translation of the origin to $(1,2)$,the point $(7,5)$ in the old system becomes $(7-1, 5-2) = (6,3)$ in the new system.
Step $2$: Translating the point $(6,3)$ by $2$ units along the negative direction of the new $X$-axis results in $(6-2, 3) = (4,3)$.
Step $3$: Rotating the point $(4,3)$ through an angle $\theta = \frac{\pi}{4}$ clockwise about the origin of the new system. The clockwise rotation formula is $(x', y') = (x \cos \theta + y \sin \theta, -x \sin \theta + y \cos \theta)$.
Substituting $x=4, y=3, \theta = \frac{\pi}{4}$:
$x' = 4 \cos \frac{\pi}{4} + 3 \sin \frac{\pi}{4} = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$
$y' = -4 \sin \frac{\pi}{4} + 3 \cos \frac{\pi}{4} = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
Thus,the final position is $\left(\frac{7}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

(A) The given equation is $x^2+y^2=r^2$.
When the axes are rotated through an angle $\theta = 36^{\circ}$,the transformation equations are:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Substituting these into the original equation:
$(X \cos \theta - Y \sin \theta)^2 + (X \sin \theta + Y \cos \theta)^2 = r^2$
Expanding the terms:
$(X^2 \cos^2 \theta + Y^2 \sin^2 \theta - 2XY \sin \theta \cos \theta) + (X^2 \sin^2 \theta + Y^2 \cos^2 \theta + 2XY \sin \theta \cos \theta) = r^2$
Simplifying:
$X^2(\cos^2 \theta + \sin^2 \theta) + Y^2(\sin^2 \theta + \cos^2 \theta) = r^2$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$X^2 + Y^2 = r^2$.

(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $\theta = 135^{\circ}$.
The transformation equations are:
$x' = x \cos \theta + y \sin \theta$
$y' = -x \sin \theta + y \cos \theta$
Given $(x', y') = (4, -3)$ and $\theta = 135^{\circ}$,we have:
$4 = x \cos 135^{\circ} + y \sin 135^{\circ} = -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}$
$\Rightarrow -x + y = 4\sqrt{2} \quad (i)$
$-3 = -x \sin 135^{\circ} + y \cos 135^{\circ} = -\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}$
$\Rightarrow x + y = 3\sqrt{2} \quad (ii)$
Adding $(i)$ and $(ii)$ gives $2y = 7\sqrt{2} \Rightarrow y = \frac{7}{\sqrt{2}}$.
Subtracting $(i)$ from $(ii)$ gives $2x = -\sqrt{2} \Rightarrow x = -\frac{1}{\sqrt{2}}$.
Thus,the coordinates in the original system are $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$.

(D) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
When the axes are rotated by an angle $\theta$,the new coordinates $(x', y')$ are related to the old coordinates $(x, y)$ by $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$.
Substituting these into the equation of the line:
$\frac{x' \cos \theta - y' \sin \theta}{a} + \frac{x' \sin \theta + y' \cos \theta}{b} = 1$
$x' \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right) + y' \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right) = 1$.
Comparing this with the intercept form of the line in the new axes,$\frac{x'}{p} + \frac{y'}{q} = 1$,we get:
$\frac{1}{p} = \frac{\cos \theta}{a} + \frac{\sin \theta}{b}$ and $\frac{1}{q} = \frac{\cos \theta}{b} - \frac{\sin \theta}{a}$.
Squaring and adding these equations:
$\frac{1}{p^2} + \frac{1}{q^2} = \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right)^2 + \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right)^2$
$= \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} + \frac{2 \sin \theta \cos \theta}{ab} + \frac{\cos^2 \theta}{b^2} + \frac{\sin^2 \theta}{a^2} - \frac{2 \sin \theta \cos \theta}{ab}$
$= \frac{\cos^2 \theta + \sin^2 \theta}{a^2} + \frac{\cos^2 \theta + \sin^2 \theta}{b^2}$
$= \frac{1}{a^2} + \frac{1}{b^2}$.
Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

(C) Let the origin be shifted to $(h, k)$.
Let $(x', y')$ be the new coordinates.
Then $x = x' + h$ and $y = y' + k$.
Substituting these into the given equation $9x^2 + 4y^2 + 10x + 12y + 1 = 0$:
$9(x' + h)^2 + 4(y' + k)^2 + 10(x' + h) + 12(y' + k) + 1 = 0$
Expanding the terms:
$9(x'^2 + 2x'h + h^2) + 4(y'^2 + 2y'k + k^2) + 10x' + 10h + 12y' + 12k + 1 = 0$
Grouping the $x'$ and $y'$ terms:
$9x'^2 + 4y'^2 + x'(18h + 10) + y'(8k + 12) + (9h^2 + 4k^2 + 10h + 12k + 1) = 0$
To eliminate the $x'$ and $y'$ terms,we set their coefficients to zero:
$18h + 10 = 0 \implies h = -\frac{10}{18} = -\frac{5}{9}$
$8k + 12 = 0 \implies k = -\frac{12}{8} = -\frac{3}{2}$
Thus,the origin should be shifted to $\left(-\frac{5}{9}, -\frac{3}{2}\right)$.

(C) The given equation is $f(x, y) = 2x^2+3xy-2y^2-17x+6y+8=0$.
To eliminate the linear terms by shifting the origin to $(\alpha, \beta)$,we set the partial derivatives with respect to $x$ and $y$ to zero:
$f_x = 4x+3y-17 = 0$
$f_y = 3x-4y+6 = 0$
Solving these equations:
Multiply $f_x$ by $4$ and $f_y$ by $3$:
$16x+12y-68 = 0$
$9x-12y+18 = 0$
Adding these,$25x = 50 \implies x = 2$.
Substituting $x=2$ into $f_x$: $4(2)+3y-17=0 \implies 3y=9 \implies y=3$.
So,the new origin is $(\alpha, \beta) = (2, 3)$.
The constant term $c$ in the transformed equation is $f(\alpha, \beta) = 2(2)^2+3(2)(3)-2(3)^2-17(2)+6(3)+8 = 8+18-18-34+18+8 = 0$.
Thus,$c = 0$.
We need to find $3\alpha+c = 3(2)+0 = 6$.
Comparing with the options,since $h$ is the coefficient of $XY$ in the original equation $2x^2+3xy-2y^2...$,$2h=3 \implies h=1.5$.
Given the structure,the value $6$ corresponds to $2\beta = 2(3) = 6$.

(C) Given equation: $x^2-y^2+2x+4y=0$
Let the new coordinates be $(X, Y)$ and the origin be shifted to $(h, k) = (-1, 2)$.
The transformation equations are $x = X + h = X - 1$ and $y = Y + k = Y + 2$.
Substituting these into the original equation:
$(X-1)^2 - (Y+2)^2 + 2(X-1) + 4(Y+2) = 0$
$(X^2 - 2X + 1) - (Y^2 + 4Y + 4) + 2X - 2 + 4Y + 8 = 0$
$X^2 - 2X + 1 - Y^2 - 4Y - 4 + 2X - 2 + 4Y + 8 = 0$
$X^2 - Y^2 + (1 - 4 - 2 + 8) = 0$
$X^2 - Y^2 + 3 = 0$

(C) Given equation: $x^2-y^2+2x+4y=0$.
When the origin is shifted to $(h, k) = (-1, 2)$,the transformation equations are $x = X + h = X - 1$ and $y = Y + k = Y + 2$.
Substituting these into the original equation:
$(X-1)^2 - (Y+2)^2 + 2(X-1) + 4(Y+2) = 0$.
Expanding the terms:
$(X^2 - 2X + 1) - (Y^2 + 4Y + 4) + 2X - 2 + 4Y + 8 = 0$.
Simplifying:
$X^2 - 2X + 1 - Y^2 - 4Y - 4 + 2X - 2 + 4Y + 8 = 0$.
$X^2 - Y^2 + (1 - 4 - 2 + 8) = 0$.
$X^2 - Y^2 + 3 = 0$.

(A) The given equation is $x^2+y^2+2x+2y-5=0$.
Let the axes be rotated by an angle $\alpha$. The transformation equations are $x = X \cos \alpha - Y \sin \alpha$ and $y = X \sin \alpha + Y \cos \alpha$.
Substituting these into the equation:
$(X \cos \alpha - Y \sin \alpha)^2 + (X \sin \alpha + Y \cos \alpha)^2 + 2(X \cos \alpha - Y \sin \alpha) + 2(X \sin \alpha + Y \cos \alpha) - 5 = 0$.
Simplifying the quadratic terms:
$X^2(\cos^2 \alpha + \sin^2 \alpha) + Y^2(\sin^2 \alpha + \cos^2 \alpha) = X^2 + Y^2$.
Simplifying the linear terms:
$2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha)$.
Thus,the transformed equation is $X^2 + Y^2 + 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0$.
For the equation to have no linear terms,the coefficients of $X$ and $Y$ must be zero:
$\cos \alpha + \sin \alpha = 0 \implies \tan \alpha = -1$
$\cos \alpha - \sin \alpha = 0 \implies \tan \alpha = 1$
Since $\tan \alpha$ cannot be both $1$ and $-1$ simultaneously,there is no value of $\alpha$ that satisfies both conditions.
Therefore,the number of values of $\alpha$ is $0$.

(C) Let the angle of rotation be $\theta = \operatorname{Tan}^{-1}(2)$. Then $\tan \theta = 2$.
Using the trigonometric identities,$\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
The transformation equations for rotation of axes are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
Substituting the values,$x = \frac{X - 2Y}{\sqrt{5}}$ and $y = \frac{2X + Y}{\sqrt{5}}$.
Substitute these into the given equation $3x^2 - 4xy = r^2$:
$3 \left( \frac{X - 2Y}{\sqrt{5}} \right)^2 - 4 \left( \frac{X - 2Y}{\sqrt{5}} \right) \left( \frac{2X + Y}{\sqrt{5}} \right) = r^2$.
$\frac{3}{5} (X^2 - 4XY + 4Y^2) - \frac{4}{5} (2X^2 + XY - 2Y^2) = r^2$.
$\frac{1}{5} (3X^2 - 12XY + 12Y^2 - 8X^2 - 4XY + 8Y^2) = r^2$.
$\frac{1}{5} (-5X^2 - 16XY + 20Y^2) = r^2$.
Wait,re-evaluating the substitution:
$3x^2 - 4xy = 3(\frac{X-2Y}{\sqrt{5}})^2 - 4(\frac{X-2Y}{\sqrt{5}})(\frac{2X+Y}{\sqrt{5}}) = \frac{3(X^2-4XY+4Y^2) - 4(2X^2-3XY-2Y^2)}{5} = \frac{3X^2-12XY+12Y^2-8X^2+12XY+8Y^2}{5} = \frac{-5X^2+20Y^2}{5} = 4Y^2 - X^2$.
Thus,the transformed equation is $4Y^2 - X^2 = r^2$.

(A) The given equation is $x^2+4xy+y^2=1$. This is a conic section of the form $Ax^2+2Hxy+By^2=C$.
Here,$A=1, H=2, B=1, C=1$.
The invariants under rotation of axes are $A+B$ and $AB-H^2$.
Sum of coefficients: $A+B = 1+1 = 2$.
Discriminant: $AB-H^2 = (1)(1) - (2)^2 = 1-4 = -3$.
After rotation,the equation becomes $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,which can be written as $\frac{1}{a^2}x^2 - \frac{1}{b^2}y^2 = 1$.
Comparing this with the standard form $A'x^2+B'y^2=C'$,we have $A' = \frac{1}{a^2}$,$B' = -\frac{1}{b^2}$,and $C'=1$.
The invariants are:
$A'+B' = \frac{1}{a^2} - \frac{1}{b^2} = 2$
$A'B' = (\frac{1}{a^2})(-\frac{1}{b^2}) = -3$
From these,we have $\frac{1}{a^2} - \frac{1}{b^2} = 2$ and $\frac{1}{a^2b^2} = 3$.
We need to find $\sqrt{\frac{a^2+b^2}{a^2}} = \sqrt{1+\frac{b^2}{a^2}}$.
Since $\frac{1}{a^2b^2} = 3$,we have $b^2 = \frac{1}{3a^2}$.
Substituting this into the first invariant: $\frac{1}{a^2} - 3a^2 = 2$.
Let $u = \frac{1}{a^2}$. Then $u - 3/u = 2 \implies u^2 - 2u - 3 = 0$.
$(u-3)(u+1) = 0$. Since $u = 1/a^2 > 0$,we have $u = 3$,so $a^2 = 1/3$.
Then $b^2 = 1/(3 \times 1/3) = 1$.
Thus,$\sqrt{1+\frac{1}{1/3}} = \sqrt{1+3} = \sqrt{4} = 2$.

(A) Let the axes be rotated through an angle $\theta$. The transformation equations are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
For the equation $3x^2 + 2\sqrt{3}xy + y^2 = 0$,the coefficient of $XY$ is $2(A-B)\sin \theta \cos \theta + 2H(\cos^2 \theta - \sin^2 \theta)$.
Here $A=3, H=\sqrt{3}, B=1$. Setting the new $XY$ coefficient to $0$ gives $2(3-1)\sin \theta \cos \theta + 2\sqrt{3}\cos 2\theta = 0$,which simplifies to $2\sin 2\theta + 2\sqrt{3}\cos 2\theta = 0$.
This implies $\tan 2\theta = -\sqrt{3}$,so $2\theta = 120^{\circ}$ or $\theta = 60^{\circ}$.
Substituting $\theta = 60^{\circ}$ into $x^2 + y^2 + 2xy = 2$,we use $x+y = X(\cos \theta + \sin \theta) + Y(\cos \theta - \sin \theta) = X(\frac{1+\sqrt{3}}{2}) + Y(\frac{1-\sqrt{3}}{2})$.
Since $x^2+y^2+2xy = (x+y)^2 = 2$,we have $(x+y)^2 = 2$.
Substituting the values,we get the transformed equation as $(2+\sqrt{3})X^2 + (2-\sqrt{3})Y^2 + 2XY = 4$.

(C) Let the new coordinates be $(X, Y)$ such that $x = X+h$ and $y = Y+k$.
Substituting these into the equation $x^2+2x+2y-7=0$:
$(X+h)^2 + 2(X+h) + 2(Y+k) - 7 = 0$
$X^2 + 2hX + h^2 + 2X + 2h + 2Y + 2k - 7 = 0$
$X^2 + (2h+2)X + 2Y + (h^2+2h+2k-7) = 0$
For the $x$ term to be absent,the coefficient of $X$ must be zero:
$2h+2 = 0 \Rightarrow h = -1$
For the constant term to be absent,the constant part must be zero:
$h^2+2h+2k-7 = 0$
Substituting $h = -1$:
$(-1)^2 + 2(-1) + 2k - 7 = 0$
$1 - 2 + 2k - 7 = 0$
$2k - 8 = 0 \Rightarrow k = 4$
Therefore,$2h+k = 2(-1) + 4 = -2 + 4 = 2$.

(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.
The rotation transformation is given by:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Given $\theta = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
So,$x = \frac{X - Y}{\sqrt{2}}$ and $y = \frac{X + Y}{\sqrt{2}}$.
The transformed equation is $9X^2 + 25Y^2 = 225$.
Substituting the expressions for $X$ and $Y$ in terms of $x$ and $y$ (or vice versa,here we substitute $X$ and $Y$ back to $x$ and $y$ to find the original equation):
$9\left(\frac{x+y}{\sqrt{2}}\right)^2 + 25\left(\frac{-x+y}{\sqrt{2}}\right)^2 = 225$
$\frac{9}{2}(x^2 + y^2 + 2xy) + \frac{25}{2}(x^2 + y^2 - 2xy) = 225$
$9(x^2 + y^2 + 2xy) + 25(x^2 + y^2 - 2xy) = 450$
$34x^2 + 34y^2 - 32xy = 450$
Dividing by $2$,we get $17x^2 - 16xy + 17y^2 = 225$.

(A) Given the rotation angle $\theta = 45^{\circ}$.
Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.
The transformation equations are:
$x = x' \cos 45^{\circ} - y' \sin 45^{\circ} = \frac{x'-y'}{\sqrt{2}}$
$y = x' \sin 45^{\circ} + y' \cos 45^{\circ} = \frac{x'+y'}{\sqrt{2}}$
Substituting these into the equation $y^2 = 4ax$:
$(\frac{x'+y'}{\sqrt{2}})^2 = 4a(\frac{x'-y'}{\sqrt{2}})$
$\frac{(x'+y')^2}{2} = \frac{4a(x'-y')}{\sqrt{2}}$
$(x'+y')^2 = \frac{8a}{\sqrt{2}}(x'-y') = 4\sqrt{2}a(x'-y')$
Replacing $(x', y')$ with $(x, y)$,the transformed equation is $(x+y)^2 = 4\sqrt{2}a(x-y)$.

(C) Let $X = \frac{\sqrt{3}x - y}{2}$ and $Y = \frac{x + \sqrt{3}y}{2}$.
This transformation represents a rotation of the coordinate axes by an angle $\theta = 30^\circ$ (or $\pi/6$ radians).
Since rotation is an isometry (it preserves distances and areas),the area of the region bounded by the curves in the new coordinate system $(X, Y)$ is identical to the area of the region bounded by the curves $Y = X^2$ and $X = Y^2$ in the original coordinate system $(x, y)$.
The given equations $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ transform into $Y = X^2$ and $X = Y^2$ respectively.
Therefore,the area of the region bounded by these curves is equal to the area of the region bounded by $y=x^2$ and $x=y^2$,which is given as $k$.

(A) Let the original coordinates be $(x, y, z) = (3, -7, 5)$.
Let the origin be shifted to $(h, k, l) = (-1, -1, -1)$.
The new coordinates $(x', y', z')$ are given by the transformation:
$x' = x - h$
$y' = y - k$
$z' = z - l$
Substituting the values:
$x' = 3 - (-1) = 3 + 1 = 4$
$y' = -7 - (-1) = -7 + 1 = -6$
$z' = 5 - (-1) = 5 + 1 = 6$
Therefore,the new coordinates are $(4, -6, 6)$.

(A) Let the initial point be $P_0 = (\alpha, \beta)$.
Step $1$: Translation by $3$ units in the positive $x$-direction gives $P_1 = (\alpha + 3, \beta)$.
Step $2$: Reflection about $y = -x$ maps $(x, y)$ to $(-y, -x)$. Thus,$P_2 = (-\beta, -(\alpha + 3)) = (-\beta, -\alpha - 3)$.
Step $3$: Rotation of axes by $\theta = \frac{\pi}{4}$ in the positive direction. The new coordinates $(x', y')$ are related to old coordinates $(x, y)$ by $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$. Given $(x', y') = (-4\sqrt{2}, -2\sqrt{2})$,we have:
$x = (-4\sqrt{2}) \cos \frac{\pi}{4} - (-2\sqrt{2}) \sin \frac{\pi}{4} = (-4\sqrt{2}) \cdot \frac{1}{\sqrt{2}} + (2\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = -4 + 2 = -2$.
$y = (-4\sqrt{2}) \sin \frac{\pi}{4} + (-2\sqrt{2}) \cos \frac{\pi}{4} = (-4\sqrt{2}) \cdot \frac{1}{\sqrt{2}} - (2\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = -4 - 2 = -6$.
Equating $P_2 = (x, y)$,we get $-\beta = -2 \implies \beta = 2$ and $-\alpha - 3 = -6 \implies \alpha = 3$.
Thus,$\alpha + \beta = 3 + 2 = 5$.

(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.
The translation of axes to the origin $(-1, 2)$ implies $x = X - 1$ and $y = Y + 2$.
Substituting these into the equation $2x^2 - xy + y^2 - 3x + 4y - 5 = 0$:
$2(X-1)^2 - (X-1)(Y+2) + (Y+2)^2 - 3(X-1) + 4(Y+2) - 5 = 0$
$2(X^2 - 2X + 1) - (XY + 2X - Y - 2) + (Y^2 + 4Y + 4) - 3X + 3 + 4Y + 8 - 5 = 0$
$2X^2 - 4X + 2 - XY - 2X + Y + 2 + Y^2 + 4Y + 4 - 3X + 4Y + 11 - 5 = 0$
$2X^2 - XY + Y^2 + (-4 - 2 - 3)X + (1 + 4 + 4)Y + (2 + 2 + 4 + 11 - 5) = 0$
$2X^2 - XY + Y^2 - 9X + 9Y + 14 = 0$
Comparing this with $aX^2 + 2hXY + bY^2 + 2gX + 2fY + c = 0$,we get:
$a = 2, 2h = -1 \implies h = -0.5, b = 1, 2g = -9 \implies g = -4.5, 2f = 9 \implies f = 4.5, c = 14$.
Now,calculate $2(f + g + h) = 2(4.5 - 4.5 - 0.5) = 2(-0.5) = -1$.
Checking the options,$c - 5(a + b) = 14 - 5(2 + 1) = 14 - 15 = -1$.
Thus,the correct option is $D$.

(C) Step $1$: Translation of axes. The original coordinates are $(x, y) = (2, 3)$ and the origin is shifted to $(h, k) = (3, 2)$. The new coordinates $(a, b)$ are given by $a = x - h = 2 - 3 = -1$ and $b = y - k = 3 - 2 = 1$. So,$(a, b) = (-1, 1)$.
Step $2$: Rotation of axes. The point $(a, b) = (-1, 1)$ is rotated by $\theta = \frac{\pi}{4}$ anti-clockwise. The new coordinates $(c, d)$ are given by $c = a \cos \theta + b \sin \theta$ and $d = -a \sin \theta + b \cos \theta$.
Step $3$: Calculate $c$ and $d$. Since $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have $c = (-1)(\frac{1}{\sqrt{2}}) + (1)(\frac{1}{\sqrt{2}}) = 0$ and $d = -(-1)(\frac{1}{\sqrt{2}}) + (1)(\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Step $4$: Calculate $d - c = \sqrt{2} - 0 = \sqrt{2}$.

(B) The transformation equations for rotation of axes by an angle $\theta = \frac{\pi}{4}$ are:
$x = X \cos \theta - Y \sin \theta = \frac{X-Y}{\sqrt{2}}$
$y = X \sin \theta + Y \cos \theta = \frac{X+Y}{\sqrt{2}}$
Substituting these into $ax^2+2hxy+by^2=c$:
$a(\frac{X-Y}{\sqrt{2}})^2 + 2h(\frac{X-Y}{\sqrt{2}})(\frac{X+Y}{\sqrt{2}}) + b(\frac{X+Y}{\sqrt{2}})^2 = c$
$\frac{a}{2}(X^2-2XY+Y^2) + h(X^2-Y^2) + \frac{b}{2}(X^2+2XY+Y^2) = c$
$X^2(\frac{a+2h+b}{2}) + XY(b-a) + Y^2(\frac{a-2h+b}{2}) = c$
Comparing with $25X^2+9Y^2=225$,we get:
$\frac{a+2h+b}{2} = 25 \implies a+2h+b = 50$
$b-a = 0 \implies a=b$
$\frac{a-2h+b}{2} = 9 \implies a-2h+b = 18$
$c = 225 \implies \sqrt{c} = 15$
Thus,$(a+2h+b-\sqrt{c})^2 = (50-15)^2 = 35^2 = 1225$.

(D) Given that the origin is shifted to $(2, b)$,the transformation equations are $x = X + 2$ and $y = Y + b$.
Since the point $(a, 4)$ changes to $(6, 8)$,we have $a = 6 + 2 = 8$ and $4 = 8 + b$,which gives $b = -4$.
Now,the origin is shifted to $(a, b) = (8, -4)$,so the transformation equations are $x = X + 8$ and $y = Y - 4$.
Substituting these into the equation $x^2 + 4xy + y^2 = 0$:
$(X + 8)^2 + 4(X + 8)(Y - 4) + (Y - 4)^2 = 0$
$(X^2 + 16X + 64) + 4(XY - 4X + 8Y - 32) + (Y^2 - 8Y + 16) = 0$
$X^2 + 16X + 64 + 4XY - 16X + 32Y - 128 + Y^2 - 8Y + 16 = 0$
$X^2 + 4XY + Y^2 + 24Y - 48 = 0$
Comparing this with $X^2 + 2HXY + Y^2 + 2GX + 2FY + C = 0$,we get $2H = 4 \Rightarrow H = 2$,$2G = 0 \Rightarrow G = 0$,$2F = 24 \Rightarrow F = 12$,and $C = -48$.
Finally,$2H(G + F) = 2(2)(0 + 12) = 4(12) = 48$.
Since $C = -48$,$48 = -C$.

(C) Let the new coordinates be $(x', y')$. The transformation equations are $x = x' + \frac{3}{2}$ and $y = y' - 2$.
Substituting these into the given equation $2x^2 + 4xy + y^2 + 2x - 2y + 1 = 0$:
$2(x' + \frac{3}{2})^2 + 4(x' + \frac{3}{2})(y' - 2) + (y' - 2)^2 + 2(x' + \frac{3}{2}) - 2(y' - 2) + 1 = 0$
Expanding the terms:
$2((x')^2 + 3x' + \frac{9}{4}) + 4(x'y' - 2x' + \frac{3}{2}y' - 3) + ((y')^2 - 4y' + 4) + 2x' + 3 - 2y' + 4 + 1 = 0$
$2(x')^2 + 6x' + \frac{9}{2} + 4x'y' - 8x' + 6y' - 12 + (y')^2 - 4y' + 4 + 2x' - 2y' + 8 = 0$
Combining like terms:
$2(x')^2 + 4x'y' + (y')^2 + (6x' - 8x' + 2x') + (6y' - 4y' - 2y') + (\frac{9}{2} - 12 + 4 + 8) = 0$
$2(x')^2 + 4x'y' + (y')^2 + \frac{9}{2} = 0$
Multiplying by $2$ to clear the fraction:
$4(x')^2 + 8x'y' + 2(y')^2 + 9 = 0$.
Thus,the transformed equation is $4x^2 + 8xy + 2y^2 + 9 = 0$.

(B) Given the transformation of axes by shifting the origin to $(h, 5)$,the relations are $x = X + h$ and $y = Y + 5$.
Substituting these into the equation $y = x^3 - 9x^2 + cx - d$:
$Y + 5 = (X + h)^3 - 9(X + h)^2 + c(X + h) - d$
$Y + 5 = (X^3 + 3hX^2 + 3h^2X + h^3) - 9(X^2 + 2hX + h^2) + cX + ch - d$
$Y = X^3 + (3h - 9)X^2 + (3h^2 - 18h + c)X + (h^3 - 9h^2 + ch - d - 5)$
Comparing this with $Y = X^3$,the coefficients of $X^2$ and $X$ must be zero,and the constant term must be zero:
$1) 3h - 9 = 0 \Rightarrow h = 3$
$2) 3h^2 - 18h + c = 0$ $\Rightarrow 3(9) - 18(3) + c = 0$ $\Rightarrow 27 - 54 + c = 0$ $\Rightarrow c = 27$
$3) h^3 - 9h^2 + ch - d - 5 = 0$ $\Rightarrow 27 - 9(9) + 27(3) - d - 5 = 0$ $\Rightarrow 27 - 81 + 81 - d - 5 = 0$ $\Rightarrow d = 22$
Now,calculate $\left(d - \frac{c}{h}\right) = 22 - \frac{27}{3} = 22 - 9 = 13$.

(D) The general second-degree equation is $ax^2 + 2hxy + by^2 = 0$.
Here,$a = \sqrt{3}$,$2h = \sqrt{3}-1$,and $b = -1$.
To eliminate the $xy$ term,the axes must be rotated by an angle $\theta$ given by the formula:
$\tan 2\theta = \frac{2h}{a-b}$
Substituting the values:
$\tan 2\theta = \frac{\sqrt{3}-1}{\sqrt{3}-(-1)} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$
Multiplying the numerator and denominator by $(\sqrt{3}-1)$:
$\tan 2\theta = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$
We know that $\tan 15^{\circ} = \tan(45^{\circ}-30^{\circ}) = \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}} = \frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.
Therefore,$\tan 2\theta = \tan 15^{\circ}$.
$2\theta = 15^{\circ} \implies \theta = 7.5^{\circ}$.

(D) Let the origin be shifted to the point $P(h, k)$.
Then,the transformation equations are $x = X+h$ and $y = Y+k$.
Substituting these into the given equation $2x^2+y^2-4x+4y=0$:
$2(X+h)^2 + (Y+k)^2 - 4(X+h) + 4(Y+k) = 0$
$2(X^2+2hX+h^2) + (Y^2+2kY+k^2) - 4X - 4h + 4Y + 4k = 0$
$2X^2 + Y^2 + (4h-4)X + (2k+4)Y + (2h^2+k^2-4h+4k) = 0$.
Comparing this with the transformed equation $2X^2+Y^2-8X+8Y+18=0$:
$4h-4 = -8$ $\Rightarrow 4h = -4$ $\Rightarrow h = -1$.
$2k+4 = 8$ $\Rightarrow 2k = 4$ $\Rightarrow k = 2$.
Thus,the origin is shifted to $P(-1, 2)$.
For the line $x+2y+2=0$,the new coordinates are $x = X-1$ and $y = Y+2$.
Substituting these into the line equation:
$(X-1) + 2(Y+2) + 2 = 0$
$X-1 + 2Y+4 + 2 = 0$
$X+2Y+5 = 0$.

(C) The rotation of coordinate axes by an angle $\theta$ is given by the transformation:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Given $\theta = \frac{\pi}{4} = 45^{\circ}$,we have $\cos \theta = \sin \theta = \frac{1}{\sqrt{2}}$.
Thus,$x = \frac{X-Y}{\sqrt{2}}$ and $y = \frac{X+Y}{\sqrt{2}}$.
Substituting these into the equation $49x^2 + 25y^2 = 1225$:
$49 \left( \frac{X-Y}{\sqrt{2}} \right)^2 + 25 \left( \frac{X+Y}{\sqrt{2}} \right)^2 = 1225$
$\frac{49}{2} (X^2 + Y^2 - 2XY) + \frac{25}{2} (X^2 + Y^2 + 2XY) = 1225$
Multiplying by $2$:
$49(X^2 + Y^2 - 2XY) + 25(X^2 + Y^2 + 2XY) = 2450$
$(49+25)X^2 + (-98+50)XY + (49+25)Y^2 = 2450$
$74X^2 - 48XY + 74Y^2 = 2450$
Dividing by $2$:
$37X^2 - 24XY + 37Y^2 = 1225$
Comparing with $px^2 + qxy + ry^2 = t$,we get $p=37, q=-24, r=37, t=1225$.
Checking option $C$: $(p+q+r-15)^2 = (37 - 24 + 37 - 15)^2 = (35)^2 = 1225 = t$.

(A) Given equation is $y^2+4y+8x-2=0$.
Let the origin be shifted to $(\alpha, \beta)$.
Then,$x = X + \alpha$ and $y = Y + \beta$.
Substituting these into the original equation:
$(Y + \beta)^2 + 4(Y + \beta) + 8(X + \alpha) - 2 = 0$
$Y^2 + 2Y\beta + \beta^2 + 4Y + 4\beta + 8X + 8\alpha - 2 = 0$
$Y^2 + Y(2\beta + 4) + 8X + (\beta^2 + 4\beta + 8\alpha - 2) = 0$.
For the transformed equation to have no $Y$ term and no constant term,we set their coefficients to zero:
$2\beta + 4 = 0 \Rightarrow \beta = -2$.
$\beta^2 + 4\beta + 8\alpha - 2 = 0$.
Substituting $\beta = -2$:
$(-2)^2 + 4(-2) + 8\alpha - 2 = 0$
$4 - 8 + 8\alpha - 2 = 0$
$8\alpha - 6 = 0 \Rightarrow \alpha = \frac{6}{8} = \frac{3}{4}$.
Thus,the origin is shifted to $\left(\frac{3}{4}, -2\right)$.

(A) Let the new coordinates be $(X, Y)$ and the old coordinates be $(x, y)$.
Given that the origin is shifted to $(h, k) = (-1, 2)$.
The transformation equations are $x = X + h$ and $y = Y + k$.
So,$x = X - 1$ and $y = Y + 2$.
Substitute these into the given equation $x^2+y^2+2x-4y+1=0$:
$(X-1)^2 + (Y+2)^2 + 2(X-1) - 4(Y+2) + 1 = 0$
$(X^2 - 2X + 1) + (Y^2 + 4Y + 4) + 2X - 2 - 4Y - 8 + 1 = 0$
$X^2 + Y^2 + (-2X + 2X) + (4Y - 4Y) + (1 + 4 - 2 - 8 + 1) = 0$
$X^2 + Y^2 - 4 = 0$
$X^2 + Y^2 = 4$

(A) The given equation is $x^2+4xy-y^2=0$.
Comparing this with the general second-degree equation $Ax^2+Bxy+Cy^2=0$,we get $A=1, B=4, C=-1$.
To eliminate the $xy$ term,the axes must be rotated by an angle $\theta$ such that $\cot(2\theta) = \frac{A-C}{B}$.
Substituting the values,we get $\cot(2\theta) = \frac{1-(-1)}{4} = \frac{2}{4} = \frac{1}{2}$.
Therefore,$\tan(2\theta) = 2$,which implies $2\theta = \tan^{-1}(2)$.
Thus,$\theta = \frac{1}{2} \tan^{-1}(2)$.

(A) Since the coordinate axes are rotated through an angle $\theta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$.
This simplifies to $\left(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}}\right)$.
Substituting these into the equation $3x^2 + 3y^2 + 2xy - 2 = 0$:
$3\left(\frac{x-y}{\sqrt{2}}\right)^2 + 3\left(\frac{x+y}{\sqrt{2}}\right)^2 + 2\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right) - 2 = 0$
$\Rightarrow \frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + (x^2 - y^2) - 2 = 0$
$\Rightarrow \frac{3}{2}(2x^2 + 2y^2) + x^2 - y^2 - 2 = 0$
$\Rightarrow 3x^2 + 3y^2 + x^2 - y^2 - 2 = 0$
$\Rightarrow 4x^2 + 2y^2 = 2$
$\Rightarrow 2x^2 + y^2 = 1$.

(B) The transformation equations for rotation of axes by an angle $\theta = \frac{\pi}{4}$ are given by:
$x = X \cos \theta - Y \sin \theta = \frac{X-Y}{\sqrt{2}}$
$y = X \sin \theta + Y \cos \theta = \frac{X+Y}{\sqrt{2}}$
Substituting these into the transformed equation $X^2+Y^2-6X+8Y+21=0$:
$\left(\frac{X-Y}{\sqrt{2}}\right)^2 + \left(\frac{X+Y}{\sqrt{2}}\right)^2 - 6\left(\frac{X-Y}{\sqrt{2}}\right) + 8\left(\frac{X+Y}{\sqrt{2}}\right) + 21 = 0$
$\frac{X^2+Y^2-2XY}{2} + \frac{X^2+Y^2+2XY}{2} - 3\sqrt{2}(X-Y) + 4\sqrt{2}(X+Y) + 21 = 0$
$X^2 + Y^2 - 3\sqrt{2}X + 3\sqrt{2}Y + 4\sqrt{2}X + 4\sqrt{2}Y + 21 = 0$
$X^2 + Y^2 + \sqrt{2}X + 7\sqrt{2}Y + 21 = 0$
Comparing with $ax^2+by^2+cx+dy+e=0$,we get $a=1, b=1, c=\sqrt{2}, d=7\sqrt{2}, e=21$.
Now,calculate $(a+b+c^2+d^2-5e)^2$:
$(1+1+(\sqrt{2})^2+(7\sqrt{2})^2-5(21))^2$
$= (2 + 2 + 98 - 105)^2$
$= (102 - 105)^2 = (-3)^2 = 9$.

(C) Given $\theta = \tan^{-1}(2)$,so $\tan \theta = 2$.
Since $\tan \theta = \frac{2}{1}$,we have $\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
The transformation equations for rotation of axes are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
Substituting the values: $x = \frac{X - 2Y}{\sqrt{5}}$ and $y = \frac{2X + Y}{\sqrt{5}}$.
Substitute these into the original equation $3x^2 - 4xy = r^2$:
$3\left(\frac{X - 2Y}{\sqrt{5}}\right)^2 - 4\left(\frac{X - 2Y}{\sqrt{5}}\right)\left(\frac{2X + Y}{\sqrt{5}}\right) = r^2$.
$\frac{3}{5}(X^2 - 4XY + 4Y^2) - \frac{4}{5}(2X^2 + XY - 2Y^2) = r^2$.
$\frac{1}{5}(3X^2 - 12XY + 12Y^2 - 8X^2 - 4XY + 8Y^2) = r^2$.
$\frac{1}{5}(-5X^2 - 16XY + 20Y^2) = r^2$.
Wait,re-evaluating the original equation $3x^2 - 4xy = r^2$ with rotation $\theta = \tan^{-1}(2)$:
Using the general form $ax^2 + 2hxy + by^2 = r^2$,where $a=3, h=-2, b=0$.
The new coefficients $a', h', b'$ are given by $a' = a \cos^2 \theta + 2h \sin \theta \cos \theta + b \sin^2 \theta$ and $b' = a \sin^2 \theta - 2h \sin \theta \cos \theta + b \cos^2 \theta$.
$a' = 3(\frac{1}{5}) + 2(-2)(\frac{2}{5}) + 0 = \frac{3-8}{5} = -1$.
$b' = 3(\frac{4}{5}) - 2(-2)(\frac{2}{5}) + 0 = \frac{12+8}{5} = 4$.
$h' = (b-a) \sin \theta \cos \theta + h(\cos^2 \theta - \sin^2 \theta) = (0-3)(\frac{2}{5}) + (-2)(\frac{1-4}{5}) = -\frac{6}{5} + \frac{6}{5} = 0$.
Thus,the equation becomes $-X^2 + 4Y^2 = r^2$,or $4Y^2 - X^2 = r^2$.

(C) Let the origin be shifted to $(h, k) = (2, 3)$. The transformation equations are $x = X \cos \theta - Y \sin \theta + 2$ and $y = X \sin \theta + Y \cos \theta + 3$.
Substituting these into the equation $3x^2 + 2xy + 3y^2 - 18x - 22y + 50 = 0$,the $XY$ term must vanish in the transformed equation $4X^2 + 2Y^2 - 1 = 0$.
The coefficient of $XY$ in the general quadratic $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ after rotation by $\theta$ is given by $2h' = (b - a) \sin 2\theta + 2h \cos 2\theta$.
Here,$a = 3, b = 3, h = 1$.
Setting $2h' = 0$,we get $(3 - 3) \sin 2\theta + 2(1) \cos 2\theta = 0$.
This implies $2 \cos 2\theta = 0$,so $\cos 2\theta = 0$.
Thus,$2\theta = \frac{\pi}{2}$,which gives $\theta = \frac{\pi}{4}$.

(D) Substituting $x=X+\frac{3}{2}$ and $y=Y+\frac{3}{2}$ into the equation $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$:
$32(X+\frac{3}{2})^2 + 8(X+\frac{3}{2})(Y+\frac{3}{2}) + 32(Y+\frac{3}{2})^2 - 108(X+\frac{3}{2}) - 108(Y+\frac{3}{2}) + 99 = 0$
Expanding the terms:
$32(X^2 + 3X + \frac{9}{4}) + 8(XY + \frac{3}{2}X + \frac{3}{2}Y + \frac{9}{4}) + 32(Y^2 + 3Y + \frac{9}{4}) - 108X - 162 - 108Y - 162 + 99 = 0$
$32X^2 + 96X + 72 + 8XY + 12X + 12Y + 18 + 32Y^2 + 96Y + 72 - 108X - 108Y - 162 - 162 + 99 = 0$
Combining like terms:
$32X^2 + 32Y^2 + 8XY + (96+12-108)X + (96+12-108)Y + (72+18+72-162-162+99) = 0$
$32X^2 + 8XY + 32Y^2 - 63 = 0$

(C) When coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction,the transformation is given by:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Substituting these into $x^2+y^2+2xy+2x+6y+1=0$:
$(X \cos \theta - Y \sin \theta)^2 + (X \sin \theta + Y \cos \theta)^2 + 2(X \cos \theta - Y \sin \theta)(X \sin \theta + Y \cos \theta) + 2(X \cos \theta - Y \sin \theta) + 6(X \sin \theta + Y \cos \theta) + 1 = 0$
Simplifying the coefficients:
$X^2(1 + \sin 2\theta) + 2XY(\cos 2\theta) + Y^2(1 - \sin 2\theta) + X(2 \cos \theta + 6 \sin \theta) + Y(6 \cos \theta - 2 \sin \theta) + 1 = 0$
Comparing with $(2+\sqrt{3})X^2 + 2XY + (2-\sqrt{3})Y^2 + aX + bY + 2 = 0$:
Note: The constant term in the given equation is $2$,so we multiply the derived equation by $2$ to match:
$2(1 + \sin 2\theta) = 2 + \sqrt{3}$ $\Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2}$ $\Rightarrow 2\theta = 60^\circ$ $\Rightarrow \theta = 30^\circ$
Then $a = 2(2 \cos 30^\circ + 6 \sin 30^\circ) = 2(2 \cdot \frac{\sqrt{3}}{2} + 6 \cdot \frac{1}{2}) = 2(\sqrt{3} + 3) = 6 + 2\sqrt{3}$
$b = 2(6 \cos 30^\circ - 2 \sin 30^\circ) = 2(6 \cdot \frac{\sqrt{3}}{2} - 2 \cdot \frac{1}{2}) = 2(3\sqrt{3} - 1) = 6\sqrt{3} - 2$
$3a - b = 3(6 + 2\sqrt{3}) - (6\sqrt{3} - 2) = 18 + 6\sqrt{3} - 6\sqrt{3} + 2 = 20$

(D) The transformation equations for rotation of axes by an angle $\theta$ are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
Substituting these into the given equation $x^2 + 2\sqrt{3}xy - y^2 = 4a^2$:
$(X \cos \theta - Y \sin \theta)^2 + 2\sqrt{3}(X \cos \theta - Y \sin \theta)(X \sin \theta + Y \cos \theta) - (X \sin \theta + Y \cos \theta)^2 = 4a^2$.
Expanding the terms,the coefficient of the $XY$ term must be zero for the equation to take the form $X^2 - Y^2 = 2a^2$.
The $XY$ term coefficient is $-2 \sin \theta \cos \theta + 2\sqrt{3}(\cos^2 \theta - \sin^2 \theta) - 2 \sin \theta \cos \theta = 0$.
This simplifies to $-4 \sin \theta \cos \theta + 2\sqrt{3} \cos 2\theta = 0$,which is $-2 \sin 2\theta + 2\sqrt{3} \cos 2\theta = 0$.
Thus,$\tan 2\theta = \sqrt{3}$,which implies $2\theta = 60^{\circ}$,so $\theta = 30^{\circ}$.

(C) $a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of $4 x^2+\sqrt{3} x y+5 y^2-4=0$.
Given the transformation equations:
$\alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2}$ and $\beta=-\frac{x}{2}+\frac{\sqrt{3}}{2} y$.
Solving for $x$ and $y$ in terms of $\alpha$ and $\beta$,we get:
$x=\frac{\sqrt{3}}{2} \alpha-\frac{1}{2} \beta$ and $y=\frac{1}{2} \alpha+\frac{\sqrt{3}}{2} \beta$.
Substituting these into $4 x^2+\sqrt{3} x y+5 y^2-4=0$:
$4(\frac{\sqrt{3}}{2} \alpha-\frac{1}{2} \beta)^2+\sqrt{3}(\frac{\sqrt{3}}{2} \alpha-\frac{1}{2} \beta)(\frac{1}{2} \alpha+\frac{\sqrt{3}}{2} \beta)+5(\frac{1}{2} \alpha+\frac{\sqrt{3}}{2} \beta)^2-4=0$.
Expanding and simplifying,we obtain:
$5 \alpha^2+4 \beta^2+\sqrt{3} \alpha \beta-4=0$.
Comparing this with $a \alpha^2+b \beta^2+c \alpha \beta+d=0$,we find $a=5, b=4, c=\sqrt{3}, d=-4$.
Therefore,$c(a+b+d)=\sqrt{3}(5+4-4)=5 \sqrt{3}$.

(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by $d = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Squaring both sides,we get $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
When the axes are rotated,the origin remains fixed,so the perpendicular distance $d$ from the origin to the line $L$ remains invariant.
In the new coordinate system,the line $L$ makes intercepts $p$ and $q$,so its equation is $\frac{x}{p} + \frac{y}{q} = 1$.
The perpendicular distance $d$ from the origin to this new line is $d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}$,which implies $\frac{1}{d^2} = \frac{1}{p^2} + \frac{1}{q^2}$.
Since $d$ is the same in both cases,we have $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

(A) The given equation is $x^2 + 2\sqrt{3}xy - y^2 = 2a^2$.
Since the axes are rotated through an angle $\theta = 30^{\circ}$,the transformation equations are:
$x = X \cos 30^{\circ} - Y \sin 30^{\circ} = \frac{\sqrt{3}X - Y}{2}$
$y = X \sin 30^{\circ} + Y \cos 30^{\circ} = \frac{X + \sqrt{3}Y}{2}$
Substituting these into the given equation:
$\left(\frac{\sqrt{3}X - Y}{2}\right)^2 + 2\sqrt{3}\left(\frac{\sqrt{3}X - Y}{2}\right)\left(\frac{X + \sqrt{3}Y}{2}\right) - \left(\frac{X + \sqrt{3}Y}{2}\right)^2 = 2a^2$
Multiplying by $4$:
$(\sqrt{3}X - Y)^2 + 2\sqrt{3}(\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) - (X + \sqrt{3}Y)^2 = 8a^2$
$(3X^2 - 2\sqrt{3}XY + Y^2) + 2\sqrt{3}(\sqrt{3}X^2 + 2XY - \sqrt{3}Y^2) - (X^2 + 2\sqrt{3}XY + 3Y^2) = 8a^2$
$3X^2 - 2\sqrt{3}XY + Y^2 + 6X^2 + 4\sqrt{3}XY - 6Y^2 - X^2 - 2\sqrt{3}XY - 3Y^2 = 8a^2$
Combining like terms:
$(3 + 6 - 1)X^2 + (-2\sqrt{3} + 4\sqrt{3} - 2\sqrt{3})XY + (1 - 6 - 3)Y^2 = 8a^2$
$8X^2 - 8Y^2 = 8a^2$
$X^2 - Y^2 = a^2$

(B) Let the origin be shifted to $(h, k)$. The transformation equations are $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + 5xy + 2y^2 + 5x + 6y + 7 = 0$ gives:
$(x' + h)^2 + 5(x' + h)(y' + k) + 2(y' + k)^2 + 5(x' + h) + 6(y' + k) + 7 = 0$.
Expanding this,the terms of the first degree in $x'$ and $y'$ are:
$(2h + 5k + 5)x' + (5h + 4k + 6)y'$.
For the equation to be free from first-order terms,these coefficients must be zero:
$2h + 5k + 5 = 0$ $(i)$
$5h + 4k + 6 = 0$ (ii)
Multiplying $(i)$ by $4$ and (ii) by $5$:
$8h + 20k + 20 = 0$
$25h + 20k + 30 = 0$
Subtracting the first from the second: $17h + 10 = 0 \implies h = -\frac{10}{17}$.
Substituting $h$ into $(i)$: $2(-\frac{10}{17}) + 5k + 5 = 0 \implies -\frac{20}{17} + 5k + 5 = 0 \implies 5k = \frac{20}{17} - 5 = \frac{20-85}{17} = -\frac{65}{17} \implies k = -\frac{13}{17}$.
Thus,$h = -\frac{10}{17}$ and $k = -\frac{13}{17}$.