Transformation of axes Questions in English

(C) We know that if the origin is shifted to $(h, k)$,then the new coordinates $(x', y')$ are given by $(x - h, y - k)$.
Given the original point $(x, y) = (4, 5)$ and the new origin $(h, k) = (1, -2)$.
Substituting these values,we get:
$x' = 4 - 1 = 3$
$y' = 5 - (-2) = 5 + 2 = 7$
Therefore,the new coordinates are $(3, 7)$.

(B) The equation of the line $L$ with respect to the original axes is $\frac{x}{a} + \frac{y}{b} = 1$.
When the axes are rotated by an angle $\alpha$,the coordinates $(x, y)$ transform to $(x', y')$ where $x = x'\cos \alpha - y'\sin \alpha$ and $y = x'\sin \alpha + y'\cos \alpha$.
Substituting these into the line equation:
$\frac{x'\cos \alpha - y'\sin \alpha}{a} + \frac{x'\sin \alpha + y'\cos \alpha}{b} = 1$
Rearranging terms:
$x'\left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right) + y'\left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right) = 1$
Comparing this to the intercept form $\frac{x'}{p} + \frac{y'}{q} = 1$,we get:
$\frac{1}{p} = \frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}$ and $\frac{1}{q} = \frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}$.
Squaring and adding these equations:
$\frac{1}{p^2} + \frac{1}{q^2} = \left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right)^2 + \left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right)^2$
$= \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha}{b^2} + \frac{2\sin \alpha \cos \alpha}{ab} + \frac{\cos^2 \alpha}{b^2} + \frac{\sin^2 \alpha}{a^2} - \frac{2\sin \alpha \cos \alpha}{ab}$
$= \frac{\cos^2 \alpha + \sin^2 \alpha}{a^2} + \frac{\sin^2 \alpha + \cos^2 \alpha}{b^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

(C) The equation of the line is $2x + y = 2$,which can be written in intercept form as $\frac{x}{1} + \frac{y}{2} = 1$. Here,the intercepts on the original axes are $a = 1$ and $b = 2$.
When the axes are rotated by an angle $\alpha = 45^\circ$ in the anticlockwise direction,the new intercepts $p$ and $q$ are given by the formulas:
$\frac{1}{p} = \frac{1}{a} \cos \alpha + \frac{1}{b} \sin \alpha$
$\frac{1}{q} = -\frac{1}{a} \sin \alpha + \frac{1}{b} \cos \alpha$
Substituting $a = 1$,$b = 2$,and $\alpha = 45^\circ$:
$\frac{1}{p} = \frac{1}{1} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{3}{2\sqrt{2}}$ $\Rightarrow p = \frac{2\sqrt{2}}{3}$
$\frac{1}{q} = -\frac{1}{1} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}}$ $\Rightarrow q = -2\sqrt{2}$
Thus,the intercepts are $\frac{2\sqrt{2}}{3}$ and $-2\sqrt{2}$.

(C) Let the original coordinates be $(x, y) = (4, -5)$ and the origin be shifted to $(h, k) = (-2, 1)$.
The new coordinates $(x', y')$ are given by the transformation formulas:
$x' = x - h = 4 - (-2) = 4 + 2 = 6$
$y' = y - k = -5 - 1 = -6$
Therefore,the new coordinates are $(6, -6)$.

(B) Let the original coordinates be $(x, y) = (4, -2\sqrt{3})$ and the angle of rotation be $\theta = 30^{\circ}$.
The transformation formulas for the new coordinates $(X, Y)$ are given by:
$X = x \cos\theta + y \sin\theta$
$Y = -x \sin\theta + y \cos\theta$
Substituting the values:
$X = 4 \cos(30^{\circ}) + (-2\sqrt{3}) \sin(30^{\circ}) = 4(\frac{\sqrt{3}}{2}) - 2\sqrt{3}(\frac{1}{2}) = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$
$Y = -4 \sin(30^{\circ}) + (-2\sqrt{3}) \cos(30^{\circ}) = -4(\frac{1}{2}) - 2\sqrt{3}(\frac{\sqrt{3}}{2}) = -2 - 3 = -5$
Thus,the new coordinates are $(\sqrt{3}, -5)$.

(A) The area of a triangle is invariant under the translation of axes.
Let the vertices of the first triangle be $A(20, 22), B(21, 24), C(22, 23)$.
By translating the origin to $(20, 22)$,the new coordinates become:
$A' = (20-20, 22-22) = (0, 0)$
$B' = (21-20, 24-22) = (1, 2)$
$C' = (22-20, 23-22) = (2, 1)$
Since the area remains invariant under translation,the area of $\triangle ABC$ is equal to the area of $\triangle A'B'C'$,which is the same as the triangle formed by points $P(0, 0), Q(1, 2),$ and $R(2, 1)$.
Thus,both Statement $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation for $(A)$.

(B) The transformation equations for rotating the axes by an angle $\theta$ are given by:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\theta = -30^{\circ}$ (negative direction),we have:
$x = x' \cos(-30^{\circ}) - y' \sin(-30^{\circ}) = x' \frac{\sqrt{3}}{2} + y' \frac{1}{2}$
$y = x' \sin(-30^{\circ}) + y' \cos(-30^{\circ}) = -x' \frac{1}{2} + y' \frac{\sqrt{3}}{2}$
Given the original point $(x, y) = (2, 1)$,we solve for $(x', y')$:
$2 = x' \frac{\sqrt{3}}{2} + y' \frac{1}{2} \implies 4 = x'\sqrt{3} + y' \quad (1)$
$1 = -x' \frac{1}{2} + y' \frac{\sqrt{3}}{2} \implies 2 = -x' + y'\sqrt{3} \quad (2)$
From $(2)$,$y'\sqrt{3} = 2 + x' \implies y' = \frac{2 + x'}{\sqrt{3}}$.
Substituting into $(1)$:
$4 = x'\sqrt{3} + \frac{2 + x'}{\sqrt{3}} \implies 4\sqrt{3} = 3x' + 2 + x' \implies 4x' = 4\sqrt{3} - 2 \implies x' = \frac{2\sqrt{3} - 1}{2}$.
Substituting $x'$ into $(2)$:
$y'\sqrt{3} = 2 + \frac{2\sqrt{3} - 1}{2} = \frac{4 + 2\sqrt{3} - 1}{2} = \frac{3 + 2\sqrt{3}}{2} \implies y' = \frac{\sqrt{3} + 2}{2}$.
Thus,the new coordinates are $\left( \frac{2\sqrt{3} - 1}{2}, \frac{2 + \sqrt{3}}{2} \right)$.

(B) Let the new coordinates be $(x', y')$. The transformation equations are given by:
$x = h + x' \cos \alpha - y' \sin \alpha$
$y = k + x' \sin \alpha + y' \cos \alpha$
Given $(x, y) = (1, 1)$,$(h, k) = (1, -2)$,and $\alpha = 30^{\circ}$.
Substituting the values:
$1 = 1 + x' \cos 30^{\circ} - y' \sin 30^{\circ} \implies x' \frac{\sqrt{3}}{2} - y' \frac{1}{2} = 0 \implies x' \sqrt{3} - y' = 0$ (Equation $1$)
$1 = -2 + x' \sin 30^{\circ} + y' \cos 30^{\circ} \implies 3 = x' \frac{1}{2} + y' \frac{\sqrt{3}}{2} \implies x' + y' \sqrt{3} = 6$ (Equation $2$)
From Equation $1$,$y' = x' \sqrt{3}$.
Substituting into Equation $2$:
$x' + (x' \sqrt{3}) \sqrt{3} = 6$
$x' + 3x' = 6 \implies 4x' = 6 \implies x' = \frac{3}{2}$.
Then $y' = \frac{3}{2} \sqrt{3} = \frac{3\sqrt{3}}{2}$.
Thus,the new coordinates are $\left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right)$.

(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y') = (4, 2)$. The angle of rotation is $\theta = -\pi / 3 = -60^{\circ}$.
The transformation equations for rotation of axes are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Substituting the values:
$x = 4 \cos(-60^{\circ}) - 2 \sin(-60^{\circ})$
$x = 4(1/2) - 2(-\sqrt{3}/2) = 2 + \sqrt{3}$
$y = 4 \sin(-60^{\circ}) + 2 \cos(-60^{\circ})$
$y = 4(-\sqrt{3}/2) + 2(1/2) = -2\sqrt{3} + 1$
Thus,the original coordinates are $(2 + \sqrt{3}, -2\sqrt{3} + 1)$.

(A) Let the new origin be $(h, k)$. The transformation equations are $x = X + h$ and $y = Y + k$.
Substituting these into the given equation: $(X + h)^2 + (Y + k)^2 - 4(X + h) + 6(Y + k) - 7 = 0$.
Expanding this: $X^2 + 2hX + h^2 + Y^2 + 2kY + k^2 - 4X - 4h + 6Y + 6k - 7 = 0$.
Grouping the terms: $X^2 + Y^2 + X(2h - 4) + Y(2k + 6) + (h^2 + k^2 - 4h + 6k - 7) = 0$.
For the first-degree terms to vanish,the coefficients of $X$ and $Y$ must be zero:
$2h - 4 = 0 \implies h = 2$.
$2k + 6 = 0 \implies k = -3$.
Thus,the origin should be shifted to $(2, -3)$.

(A) The given point is $P(2, -1)$.
The reflection of point $P(2, -1)$ about the $y$-axis is $P'(-2, -1)$.
When the axes are rotated by an angle $\theta$ in the negative direction (clockwise),the rotation angle is $\theta = -45^{\circ}$.
The transformation equations for rotation of axes are:
$X = x \cos \theta + y \sin \theta$
$Y = -x \sin \theta + y \cos \theta$
Substituting $x = -2$,$y = -1$,and $\theta = -45^{\circ}$:
$X = (-2) \cos(-45^{\circ}) + (-1) \sin(-45^{\circ}) = (-2) \left( \frac{1}{\sqrt{2}} \right) + (-1) \left( -\frac{1}{\sqrt{2}} \right) = -\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
$Y = -(-2) \sin(-45^{\circ}) + (-1) \cos(-45^{\circ}) = 2 \left( -\frac{1}{\sqrt{2}} \right) - 1 \left( \frac{1}{\sqrt{2}} \right) = -\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$
Thus,the new coordinates are $\left( -\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}} \right)$.

(B) Given equation: $x^2 - 3xy + 11x - 12y + 36 = 0$.
Let the new coordinates be $(X, Y)$ such that $x = X - 4$ and $y = Y + 1$.
Substituting these into the equation:
$(X - 4)^2 - 3(X - 4)(Y + 1) + 11(X - 4) - 12(Y + 1) + 36 = 0$.
Expanding the terms:
$(X^2 - 8X + 16) - 3(XY + X - 4Y - 4) + 11X - 44 - 12Y - 12 + 36 = 0$.
$X^2 - 8X + 16 - 3XY - 3X + 12Y + 12 + 11X - 12Y - 20 = 0$.
Simplifying:
$X^2 - 3XY + 8 = 0$.
Dividing by $8$:
$\frac{X^2}{8} - \frac{3}{8}XY + 1 = 0$.
Comparing with $ax^2 + bxy + 1 = 0$,we get $a = \frac{1}{8}$ and $b = -\frac{3}{8}$.
Therefore,$b^2 - a = (-\frac{3}{8})^2 - \frac{1}{8} = \frac{9}{64} - \frac{8}{64} = \frac{1}{64}$.

(C) Let the original coordinates be $(x, y) = (-5, 3)$.
Let the origin be shifted to $(h, k) = (2, -5)$.
The transformation formulas for the new coordinates $(x', y')$ are given by:
$x' = x - h$
$y' = y - k$
Substituting the given values:
$x' = -5 - 2 = -7$
$y' = 3 - (-5) = 3 + 5 = 8$
Thus,the new coordinates are $(-7, 8)$.

(C) Step $1$: Reflection of $(4, 1)$ about the line $y = x$ gives the point $(1, 4)$.
Step $2$: Translation of $(1, 4)$ by $2$ units along the positive $x$-axis gives $(1 + 2, 4) = (3, 4)$.
Step $3$: Rotation of $(3, 4)$ through an angle $\theta = \pi/4$ about the origin in the anti-clockwise direction is given by the transformation:
$x' = x \cos \theta - y \sin \theta = 3 \cos(\pi/4) - 4 \sin(\pi/4) = 3(1/\sqrt{2}) - 4(1/\sqrt{2}) = -1/\sqrt{2}$
$y' = x \sin \theta + y \cos \theta = 3 \sin(\pi/4) + 4 \cos(\pi/4) = 3(1/\sqrt{2}) + 4(1/\sqrt{2}) = 7/\sqrt{2}$
Thus,the final position is $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$.

(D) Let $(x', y')$ be the coordinates on the new axes. The transformation equations are $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$.
Substituting these into the general equation $ax^2 + by^2 + 2hxy + 2fy + 2gx + c = 0$,the coefficient of the $x'y'$ term must be zero for the mixed term to be removed.
The coefficient of $x'y'$ is given by $2(b - a) \sin \theta \cos \theta + 2h(\cos^2 \theta - \sin^2 \theta) = 0$.
This simplifies to $-(a - b) \sin 2\theta + 2h \cos 2\theta = 0$.
Rearranging gives $2h \cos 2\theta = (a - b) \sin 2\theta$.
Therefore,$\tan 2\theta = \frac{2h}{a - b}$.

(A) When the axes are rotated by an angle $\theta$ in the clockwise direction,the new coordinates $(x', y')$ of a point $(x, y)$ are given by the transformation equations:
$x' = x \cos \theta + y \sin \theta$
$y' = -x \sin \theta + y \cos \theta$
Given $x = 4$,$y = 2$,and $\theta = \frac{\pi}{3}$,we have $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
Substituting these values:
$x' = 4 \left(\frac{1}{2}\right) + 2 \left(\frac{\sqrt{3}}{2}\right) = 2 + \sqrt{3}$
$y' = -4 \left(\frac{\sqrt{3}}{2}\right) + 2 \left(\frac{1}{2}\right) = -2\sqrt{3} + 1$
Thus,the new coordinates are $(2 + \sqrt{3}, -2\sqrt{3} + 1)$.

(D) Let the original unit vectors be $\hat{i}, \hat{j}, \hat{k}$. When the coordinate system is rotated by $\pi / 2$ about the $z-$axis such that the $x-$axis moves to the $y-$axis,the new unit vectors $\hat{i}'$ and $\hat{j}'$ are related to the old ones as follows:
$\hat{i}' = \hat{j}$
$\hat{j}' = -\hat{i}$
$\hat{k}' = \hat{k}$
Given $\vec{a} = 2\hat{i} + 3\hat{j} + 7\hat{k}$.
We can express $\hat{i} = -\hat{j}'$ and $\hat{j} = \hat{i}'$.
Substituting these into the expression for $\vec{a}$:
$\vec{a} = 2(-\hat{j}') + 3(\hat{i}') + 7\hat{k}'$
$\vec{a} = 3\hat{i}' - 2\hat{j}' + 7\hat{k}'$
Thus,the new components are $(3, -2, 7)$.

(D) The magnitude of a vector remains invariant under the rotation of the coordinate axes.
Given $\vec{a}_{Old} = (3p, 1)$ and $\vec{a}_{New} = (p+1, \sqrt{10})$.
Equating the squares of the magnitudes:
$|\vec{a}_{Old}|^2 = |\vec{a}_{New}|^2$
$(3p)^2 + 1^2 = (p+1)^2 + (\sqrt{10})^2$
$9p^2 + 1 = p^2 + 2p + 1 + 10$
$8p^2 - 2p - 10 = 0$
$4p^2 - p - 5 = 0$
$(4p - 5)(p + 1) = 0$
Thus,$p = \frac{5}{4}$ or $p = -1$.
Comparing with the given options,the correct value is $-1$.

(B) Let the origin be shifted to $(\alpha, \beta)$. The transformation equations are $x = X + \alpha$ and $y = Y + \beta$.
Substituting these into the original equation $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$:
$3(X + \alpha)^2 + 4(Y + \beta)^2 - (X + \alpha)(Y + \beta) - 5(X + \alpha) - 7(Y + \beta) + 2 = 0$
Expanding this,the coefficients of $X$ and $Y$ must be zero for the equation to be of the form $3X^2 + 4Y^2 - XY + k = 0$.
The coefficient of $X$ is $6\alpha - \beta - 5 = 0$.
The coefficient of $Y$ is $8\beta - \alpha - 7 = 0$.
Solving these linear equations: $\alpha = 1, \beta = 1$.
Substituting $\alpha = 1, \beta = 1$ into the constant term:
$k = 3(1)^2 + 4(1)^2 - (1)(1) - 5(1) - 7(1) + 2 = 3 + 4 - 1 - 5 - 7 + 2 = -4$.
Thus,$\alpha + \beta - k = 1 + 1 - (-4) = 6$.

(A) The original equation is $x^2+y^2=4$.
Rotation of axes about the origin does not change the form of the equation $x^2+y^2=r^2$ because the distance from the origin remains invariant.
Let the coordinates after rotation be $(x', y')$. Then $x'^2+y'^2=4$.
Next,the axes are translated to the new origin $(h, k) = (2, -2)$.
The transformation equations are $x' = X + 2$ and $y' = Y - 2$.
Substituting these into the equation $x'^2+y'^2=4$:
$(X+2)^2 + (Y-2)^2 = 4$
$X^2 + 4X + 4 + Y^2 - 4Y + 4 = 4$
$X^2 + Y^2 + 4X - 4Y + 4 = 0$.
Comparing this with $X^2+Y^2+aX+bY+c=0$,we get $a=4$,$b=-4$,and $c=4$.
Therefore,$a+b+c = 4 - 4 + 4 = 4$.

(C) Let the initial coordinates be $(x, y) = (4, 1)$.
$(i)$ After shifting the origin to $(1, 6)$,the new coordinates $(x', y')$ are given by $x' = x - 1 = 4 - 1 = 3$ and $y' = y - 6 = 1 - 6 = -5$. So,$P' = (3, -5)$.
(ii) After translation by $2$ units along the positive $X$-axis,the new coordinates $(x'', y'')$ are $x'' = x' + 2 = 3 + 2 = 5$ and $y'' = y' = -5$. So,$P'' = (5, -5)$.
(iii) After rotation of axes by $90^{\circ}$ in the positive direction,the new coordinates $(X, Y)$ are given by $X = x'' \cos(90^{\circ}) + y'' \sin(90^{\circ}) = 5(0) + (-5)(1) = -5$ and $Y = -x'' \sin(90^{\circ}) + y'' \cos(90^{\circ}) = -5(1) + (-5)(0) = -5$.
Thus,the final coordinates are $(-5, -5)$.

(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.
The transformation equations are $x = X + h$ and $y = Y + k$,where $(h, k) = (2, 3)$.
So,$x = X + 2$ and $y = Y + 3$.
Substituting these into the original equation $x^2 + 3xy - 2y^2 + 4x - y - 20 = 0$:
$(X+2)^2 + 3(X+2)(Y+3) - 2(Y+3)^2 + 4(X+2) - (Y+3) - 20 = 0$
$(X^2 + 4X + 4) + 3(XY + 3X + 2Y + 6) - 2(Y^2 + 6Y + 9) + 4X + 8 - Y - 3 - 20 = 0$
$X^2 + 4X + 4 + 3XY + 9X + 6Y + 18 - 2Y^2 - 12Y - 18 + 4X - Y - 3 - 20 = 0$
Grouping the terms:
$X^2 + 3XY - 2Y^2 + (4 + 9 + 4)X + (6 - 12 - 1)Y + (4 + 18 - 18 - 3 - 20) = 0$
$X^2 + 3XY - 2Y^2 + 17X - 7Y - 19 = 0$
Comparing this with $AX^2 + BXY + CY^2 + DX + EY + F = 0$,we get $D = 17$,$E = -7$,and $F = -19$.
Therefore,$D + E + F = 17 - 7 - 19 = -9$.
Wait,re-calculating the constant term: $4 + 18 - 18 - 3 - 20 = 4 - 23 = -19$.
Re-calculating $D$: $4 + 9 + 4 = 17$.
Re-calculating $E$: $6 - 12 - 1 = -7$.
Sum $D+E+F = 17 - 7 - 19 = -9$. Since $-9$ is not in the options,let us re-check the calculation.
$x^2+3xy-2y^2+4x-y-20=0$ at $(2,3)$:
$x^2+4x+4 + 3(xy+3x+2y+6) - 2(y^2+6y+9) + 4x+8 - y-3 - 20 = 0$
$x^2+3xy-2y^2 + (4+9+4)x + (6-12-1)y + (4+18-18+8-3-20) = 0$
$x^2+3xy-2y^2 + 17x - 7y - 11 = 0$.
$D+E+F = 17 - 7 - 11 = -1$.

(A) The vertices of the triangle are $A(7,5), B(-5,-7), C(7,-7)$.
Since $AC$ is vertical $(x=7)$ and $BC$ is horizontal $(y=-7)$,the triangle is a right-angled triangle at $C(7,-7)$.
The orthocentre $H$ of a right-angled triangle is the vertex at the right angle,so $H = (7,-7)$.
The axes are translated to $(7,-7)$,so the new coordinates $(X, Y)$ are related to the old coordinates $(x, y)$ by $X = x - 7$ and $Y = y + 7$.
The side lengths are $a = BC = 12$,$b = AC = 12$,and $c = AB = \sqrt{(7 - (-5))^2 + (5 - (-7))^2} = \sqrt{12^2 + 12^2} = 12\sqrt{2}$.
The incentre $I(x_i, y_i)$ is given by $(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c})$.
$x_i = \frac{12(7) + 12(-5) + 12\sqrt{2}(7)}{12 + 12 + 12\sqrt{2}} = \frac{24 + 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{2 + 7\sqrt{2}}{2 + \sqrt{2}} = \frac{(2 + 7\sqrt{2})(2 - \sqrt{2})}{4 - 2} = \frac{4 - 2\sqrt{2} + 14\sqrt{2} - 14}{2} = \frac{12\sqrt{2} - 10}{2} = 6\sqrt{2} - 5$.
$y_i = \frac{12(5) + 12(-7) + 12\sqrt{2}(-7)}{12 + 12 + 12\sqrt{2}} = \frac{60 - 84 - 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{-24 - 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{-2 - 7\sqrt{2}}{2 + \sqrt{2}} = \frac{(-2 - 7\sqrt{2})(2 - \sqrt{2})}{4 - 2} = \frac{-4 + 2\sqrt{2} - 14\sqrt{2} + 14}{2} = \frac{10 - 12\sqrt{2}}{2} = 5 - 6\sqrt{2}$.
In the new system,$X_i = x_i - 7 = 6\sqrt{2} - 5 - 7 = 6\sqrt{2} - 12$ and $Y_i = y_i - (-7) = 5 - 6\sqrt{2} + 7 = 12 - 6\sqrt{2}$.

(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The rotation transformation is given by:
$x = x' \cos 60^{\circ} - y' \sin 60^{\circ} = \frac{x'}{2} - \frac{\sqrt{3}y'}{2}$
$y = x' \sin 60^{\circ} + y' \cos 60^{\circ} = \frac{\sqrt{3}x'}{2} + \frac{y'}{2}$
Substituting these into the equation $x+y=1$:
$(\frac{x'}{2} - \frac{\sqrt{3}y'}{2}) + (\frac{\sqrt{3}x'}{2} + \frac{y'}{2}) = 1$
$x'(\frac{1+\sqrt{3}}{2}) + y'(\frac{1-\sqrt{3}}{2}) = 1$
This is in the form $\frac{x'}{a} + \frac{y'}{b} = 1$,where $a = \frac{2}{1+\sqrt{3}}$ and $b = \frac{2}{1-\sqrt{3}}$.
Then $\frac{1}{a} = \frac{1+\sqrt{3}}{2}$ and $\frac{1}{b} = \frac{1-\sqrt{3}}{2}$.
$\frac{1}{a^2} + \frac{1}{b^2} = (\frac{1+\sqrt{3}}{2})^2 + (\frac{1-\sqrt{3}}{2})^2$
$= \frac{1+3+2\sqrt{3}}{4} + \frac{1+3-2\sqrt{3}}{4} = \frac{4+2\sqrt{3} + 4-2\sqrt{3}}{4} = \frac{8}{4} = 2$.

(C) Step $1$: Translation of axes.
Given original point $(x, y) = (-1, 2)$ and new origin $(h, k) = (2, -1)$.
The new coordinates $(a, b)$ are given by $a = x - h = -1 - 2 = -3$ and $b = y - k = 2 - (-1) = 3$.
So,$(a, b) = (-3, 3)$.
Step $2$: Rotation of axes.
The axes are rotated by $\theta = 45^{\circ}$ about the new origin $(2, -1)$.
The new coordinates $(c, d)$ are given by:
$c = a \cos \theta + b \sin \theta = -3 \cos 45^{\circ} + 3 \sin 45^{\circ} = -3(\frac{1}{\sqrt{2}}) + 3(\frac{1}{\sqrt{2}}) = 0$.
$d = -a \sin \theta + b \cos \theta = -(-3) \sin 45^{\circ} + 3 \cos 45^{\circ} = 3(\frac{1}{\sqrt{2}}) + 3(\frac{1}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
So,$(c, d) = (0, 3\sqrt{2})$.
Step $3$: Reflection about $y = x$.
The reflection of a point $(x, y)$ about the line $y = x$ is $(y, x)$.
Therefore,the reflection of $(0, 3\sqrt{2})$ is $(3\sqrt{2}, 0)$.
Thus,$(e, f) = (3\sqrt{2}, 0)$.

(B) Given equation: $x^2-y^2+2y-1=0$.
Let the origin be shifted to $(0, k)$ such that $x=X$ and $y=Y+k$.
Substituting these into the equation:
$X^2-(Y+k)^2+2(Y+k)-1=0$
$X^2-(Y^2+2kY+k^2)+2Y+2k-1=0$
$X^2-Y^2+Y(2-2k)-k^2+2k-1=0$.
To remove the $Y$-term,set the coefficient of $Y$ to zero:
$2-2k=0 \Rightarrow k=1$.
Substituting $k=1$ into the equation:
$X^2-Y^2-(1)^2+2(1)-1=0$
$X^2-Y^2-1+2-1=0$
$X^2-Y^2=0$.
Thus,the transformed equation is $x^2-y^2=0$.

(D) Given equation: $2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0$.
Let the origin be shifted to $(h, k)$,so $x = X + h$ and $y = Y + k$.
Substituting these into the equation: $2(X+h)^2 - 3(Y+k)^2 + 4(X+h)(Y+k) + 4(X+h) + 4(Y+k) - 14 = 0$.
Expanding the terms: $2X^2 - 3Y^2 + 4XY + (4h + 4k + 4)X + (4h - 6k + 4)Y + (2h^2 - 3k^2 + 4hk + 4h + 4k - 14) = 0$.
To remove the first-degree terms,set the coefficients of $X$ and $Y$ to zero:
$4h + 4k + 4 = 0 \Rightarrow h + k = -1$
$4h - 6k + 4 = 0 \Rightarrow 2h - 3k = -2$
Solving these equations: $h = -1, k = 0$.
Thus,the transformation is $x = X - 1$ and $y = Y$.
Now,substitute these into $x^2 + y^2 - 3xy + 4y + 3 = 0$:
$(X - 1)^2 + Y^2 - 3(X - 1)Y + 4Y + 3 = 0$
$X^2 - 2X + 1 + Y^2 - 3XY + 3Y + 4Y + 3 = 0$
$X^2 + Y^2 - 3XY - 2X + 7Y + 4 = 0$.

(A) Given equation: $3x^2+2xy+3y^2-18x-22y+50=0$.
Shifting origin to $(2,3)$,substitute $x=X+2, y=Y+3$:
$3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2-18(X+2)-22(Y+3)+50=0$.
Simplifying,we get $3X^2+2XY+3Y^2-1=0$.
Now,rotating axes by angle $\theta$,substitute $X=x'\cos\theta-y'\sin\theta$ and $Y=x'\sin\theta+y'\cos\theta$:
$3(x'\cos\theta-y'\sin\theta)^2+2(x'\cos\theta-y'\sin\theta)(x'\sin\theta+y'\cos\theta)+3(x'\sin\theta+y'\cos\theta)^2-1=0$.
Expanding and collecting terms:
$(3\cos^2\theta+3\sin^2\theta+2\sin\theta\cos\theta)x'^2 + (3\sin^2\theta+3\cos^2\theta-2\sin\theta\cos\theta)y'^2 + (2\cos^2\theta-2\sin^2\theta)x'y' - 1 = 0$.
$(3+\sin 2\theta)x'^2 + (3-\sin 2\theta)y'^2 + (2\cos 2\theta)x'y' - 1 = 0$.
Comparing with $4x^2+2y^2-1=0$,the coefficient of $x'y'$ must be $0$:
$2\cos 2\theta = 0$ $\Rightarrow 2\theta = \frac{\pi}{2}$ $\Rightarrow \theta = \frac{\pi}{4}$.

(B) When the coordinate axes are rotated through an angle $\theta$,the new coordinates $(x', y')$ of a point $(x, y)$ are given by the transformation equations:
$x' = x \cos \theta + y \sin \theta$
$y' = -x \sin \theta + y \cos \theta$
Given $(x, y) = (2 \sqrt{2}, -3 \sqrt{2})$ and $\theta = 45^{\circ}$.
Substituting the values:
$x' = (2 \sqrt{2}) \cos 45^{\circ} + (-3 \sqrt{2}) \sin 45^{\circ}$
$x' = (2 \sqrt{2}) \times \frac{1}{\sqrt{2}} - (3 \sqrt{2}) \times \frac{1}{\sqrt{2}} = 2 - 3 = -1$
$y' = -(2 \sqrt{2}) \sin 45^{\circ} + (-3 \sqrt{2}) \cos 45^{\circ}$
$y' = -(2 \sqrt{2}) \times \frac{1}{\sqrt{2}} - (3 \sqrt{2}) \times \frac{1}{\sqrt{2}} = -2 - 3 = -5$
Thus,the new coordinates are $(-1, -5)$.

(B) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y') = (4, -3)$ after rotation by $\theta = 135^{\circ}$.
Using the transformation formulas for rotation of axes:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\cos 135^{\circ} = -\frac{1}{\sqrt{2}}$ and $\sin 135^{\circ} = \frac{1}{\sqrt{2}}$.
Substituting the values:
$x = 4 \left(-\frac{1}{\sqrt{2}}\right) - (-3) \left(\frac{1}{\sqrt{2}}\right) = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
$y = 4 \left(\frac{1}{\sqrt{2}}\right) + (-3) \left(-\frac{1}{\sqrt{2}}\right) = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$
Thus,the original coordinates are $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

(B) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y) = (1, -1)$ after rotation by $\theta = 45^{\circ}$.
Using the transformation formulas:
$x = X \cos \theta - Y \sin \theta$
$x = (1) \cos 45^{\circ} - (-1) \sin 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
$y = X \sin \theta + Y \cos \theta$
$y = (1) \sin 45^{\circ} + (-1) \cos 45^{\circ} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$
Therefore,the coordinates of $P$ in the original system are $(\sqrt{2}, 0)$.

(B) Given equation is $3x^2 + 3y^2 + 2xy = 2 \dots (i)$.
When coordinate axes are rotated through an angle $\theta = 45^{\circ}$,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 equation $(i)$:
$3\left(\frac{X - Y}{\sqrt{2}}\right)^2 + 3\left(\frac{X + Y}{\sqrt{2}}\right)^2 + 2\left(\frac{X - Y}{\sqrt{2}}\right)\left(\frac{X + Y}{\sqrt{2}}\right) = 2$
$\frac{3}{2}(X^2 - 2XY + Y^2) + \frac{3}{2}(X^2 + 2XY + Y^2) + (X^2 - Y^2) = 2$
$\frac{3}{2}(2X^2 + 2Y^2) + X^2 - Y^2 = 2$
$3X^2 + 3Y^2 + X^2 - Y^2 = 2$
$4X^2 + 2Y^2 = 2$
Dividing by $2$,we get $2X^2 + Y^2 = 1$.
Thus,the transformed equation is $2x^2 + y^2 = 1$.

(D) The given equation is $x^2+2 \sqrt{3} xy - y^2 = 8$.
When the axes are rotated through an angle $\theta = \frac{\pi}{3}$,the coordinates $(x, y)$ are replaced by $(X, Y)$ where:
$x = X \cos \frac{\pi}{3} - Y \sin \frac{\pi}{3} = \frac{X - \sqrt{3}Y}{2}$
$y = X \sin \frac{\pi}{3} + Y \cos \frac{\pi}{3} = \frac{\sqrt{3}X + Y}{2}$
Substituting these into the equation:
$\left(\frac{X - \sqrt{3}Y}{2}\right)^2 + 2 \sqrt{3} \left(\frac{X - \sqrt{3}Y}{2}\right) \left(\frac{\sqrt{3}X + Y}{2}\right) - \left(\frac{\sqrt{3}X + Y}{2}\right)^2 = 8$
$\frac{1}{4} [ (X^2 - 2\sqrt{3}XY + 3Y^2) + 2\sqrt{3}(\sqrt{3}X^2 + XY - 3XY - \sqrt{3}Y^2) - (3X^2 + 2\sqrt{3}XY + Y^2) ] = 8$
$\frac{1}{4} [ X^2 - 2\sqrt{3}XY + 3Y^2 + 6X^2 - 4\sqrt{3}XY - 6Y^2 - 3X^2 - 2\sqrt{3}XY - Y^2 ] = 8$
$\frac{1}{4} [ 4X^2 - 8\sqrt{3}XY - 4Y^2 ] = 8$
$X^2 - 2\sqrt{3}XY - Y^2 = 8$
Thus,the transformed equation is $x^2 - 2\sqrt{3}xy - y^2 = 8$.

(B) When the origin is shifted to the point $(h, k) = (-1, 1)$,the new coordinates $(X, Y)$ are related to the old coordinates $(x, y)$ by the equations:
$x = X + h = X - 1$
$y = Y + k = Y + 1$
Substituting these into the given equation $3x^2 + y^2 + 2x + 4y + 15 = 0$:
$3(X - 1)^2 + (Y + 1)^2 + 2(X - 1) + 4(Y + 1) + 15 = 0$
$3(X^2 - 2X + 1) + (Y^2 + 2Y + 1) + 2X - 2 + 4Y + 4 + 15 = 0$
$3X^2 - 6X + 3 + Y^2 + 2Y + 1 + 2X - 2 + 4Y + 4 + 15 = 0$
$3X^2 + Y^2 - 4X + 6Y + 21 = 0$
Thus,the transformed equation is $3x^2 + y^2 - 4x + 6y + 21 = 0$.

(A) Given the equation of the line is $x \cos \theta + y \sin \theta = p$ ...$(i)$
When the axes are rotated through an angle $\theta$,the transformation equations are:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Substituting these values into equation $(i)$:
$(X \cos \theta - Y \sin \theta) \cos \theta + (X \sin \theta + Y \cos \theta) \sin \theta = p$
$X \cos^2 \theta - Y \sin \theta \cos \theta + X \sin^2 \theta + Y \sin \theta \cos \theta = p$
$X (\cos^2 \theta + \sin^2 \theta) = p$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$X = p$

(B) Let $(x, y)$ be the coordinates in the old system and $(X, Y)$ be the coordinates in the new system after rotation by an angle $\theta = 45^{\circ}$.
The transformation equations are:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Given $x = 4 \sqrt{2}$,$y = -6 \sqrt{2}$,and $\theta = 45^{\circ}$,we have $\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Substituting these values:
$4 \sqrt{2} = X \left(\frac{1}{\sqrt{2}}\right) - Y \left(\frac{1}{\sqrt{2}}\right) \Rightarrow X - Y = 8$
$-6 \sqrt{2} = X \left(\frac{1}{\sqrt{2}}\right) + Y \left(\frac{1}{\sqrt{2}}\right) \Rightarrow X + Y = -12$
Adding the two equations: $2X = -4 \Rightarrow X = -2$.
Subtracting the first from the second: $2Y = -20 \Rightarrow Y = -10$.
Thus,the new coordinates are $(-2, -10)$.

(B) When the origin is shifted to $(h, k) = (2, 3)$,the relation between the old coordinates $(X, Y)$ and new coordinates $(x, y)$ is $X = x + h = x + 2$ and $Y = y + k = y + 3$.
To find the original equation,we substitute $x$ with $(X-2)$ and $y$ with $(Y-3)$ in the given transformed equation:
$(X-2)^2 + 3(X-2)(Y-3) - 2(Y-3)^2 + 17(X-2) - 7(Y-3) - 11 = 0$
Expanding the terms:
$(X^2 - 4X + 4) + 3(XY - 3X - 2Y + 6) - 2(Y^2 - 6Y + 9) + 17X - 34 - 7Y + 21 - 11 = 0$
$X^2 - 4X + 4 + 3XY - 9X - 6Y + 18 - 2Y^2 + 12Y - 18 + 17X - 7Y - 24 = 0$
Grouping the terms:
$X^2 + 3XY - 2Y^2 + (-4 - 9 + 17)X + (-6 + 12 - 7)Y + (4 + 18 - 18 - 24) = 0$
$X^2 + 3XY - 2Y^2 + 4X - Y - 20 = 0$
Thus,the original equation is $x^2+3xy-2y^2+4x-y-20=0$.

(C) The given equation is $y^2-6y-4x+13=0$.
Completing the square for the $y$ terms:
$(y^2-6y+9)-9-4x+13=0$
$(y-3)^2-4x+4=0$
$(y-3)^2-4(x-1)=0$.
Let the new origin be $(h,k)$. We substitute $y = Y+k$ and $x = X+h$.
To transform the equation to the form $Y^2+AX=0$,we set $k=3$ and $h=1$.
Thus,the origin should be shifted to $(1,3)$.
Therefore,option $C$ is correct.

(B) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y') = (4, -3)$. The rotation angle is $\theta = 135^{\circ}$.
The transformation formulas for rotation of axes are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\cos 135^{\circ} = -\frac{1}{\sqrt{2}}$ and $\sin 135^{\circ} = \frac{1}{\sqrt{2}}$,we substitute these values:
$x = 4 \left(-\frac{1}{\sqrt{2}}\right) - (-3) \left(\frac{1}{\sqrt{2}}\right)$
$x = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
$y = 4 \left(\frac{1}{\sqrt{2}}\right) + (-3) \left(-\frac{1}{\sqrt{2}}\right)$
$y = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$
Thus,the original coordinates are $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

(D) Given equation is $y^2-6y-4x+13=0$.
Let the origin be shifted to $(h, k)$.
Then $y$ becomes $y+k$ and $x$ becomes $x+h$.
Substituting these in the equation: $(y+k)^2 - 6(y+k) - 4(x+h) + 13 = 0$.
Expanding this: $y^2 + 2ky + k^2 - 6y - 6k - 4x - 4h + 13 = 0$.
Grouping terms: $y^2 + (2k-6)y - 4x + (k^2 - 6k - 4h + 13) = 0$.
For the equation to have no term in $y$,the coefficient of $y$ must be zero: $2k - 6 = 0 \implies k = 3$.
For the equation to have no constant term,the constant part must be zero: $k^2 - 6k - 4h + 13 = 0$.
Substituting $k = 3$: $(3)^2 - 6(3) - 4h + 13 = 0$.
$9 - 18 - 4h + 13 = 0$.
$-9 - 4h + 13 = 0$.
$4 - 4h = 0 \implies h = 1$.
Thus,the origin should be shifted to $(1, 3)$.

(C) $1$) The area of a triangle depends on the coordinates of its vertices. Since translation of axes shifts the origin but keeps the axes parallel,the relative distances and coordinates differences remain unchanged. Thus,the area is invariant. This statement is true.
$2$) The slope of a line is defined by the ratio of the change in $y$ to the change in $x$. Under translation,$x = X + h$ and $y = Y + k$,so $dx = dX$ and $dy = dY$. The slope $m = \frac{dy}{dx} = \frac{dY}{dX}$ remains invariant. This statement is true.
$3$) By definition,the translation of axes involves shifting the origin to a new point $(h, k)$ while keeping the axes parallel to the original axes. If the direction of the axes is changed,it is called rotation of axes. Therefore,this statement is false.
$4$) If the origin is shifted to $(h, k)$,the new coordinates $(X, Y)$ are related to the old coordinates $(x, y)$ by $x = X + h$ and $y = Y + k$. Substituting these into the original equation $F(x, y) = 0$ gives $F(X+h, Y+k) = 0$. The statement provided is the inverse of this transformation logic. This statement is false.
Note: In standard coordinate geometry,statement $3$ is false because translation does not involve changing the direction of the axes.

(B) When the coordinate axes are rotated by an angle $\theta = \frac{\pi}{4}$ in the positive direction,the transformation equations 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 the equation $25x^2 + 9y^2 = 225$:
$25 \left( \frac{X-Y}{\sqrt{2}} \right)^2 + 9 \left( \frac{X+Y}{\sqrt{2}} \right)^2 = 225$
$\frac{25}{2} (X^2 + Y^2 - 2XY) + \frac{9}{2} (X^2 + Y^2 + 2XY) = 225$
$\frac{34X^2 + 34Y^2 - 32XY}{2} = 225$
$17X^2 - 16XY + 17Y^2 = 225$
Comparing this with $\alpha x^2 + \beta xy + \gamma y^2 = \delta$,we get $\alpha = 17$,$\beta = -16$,$\gamma = 17$,and $\delta = 225$.
Now,calculate $(\alpha + \beta + \gamma - \sqrt{\delta})^2$:
$(17 - 16 + 17 - \sqrt{225})^2 = (18 - 15)^2 = 3^2 = 9$.

(B) The given equation is $2 x^2+2 y^2-8 x-12 y+18=0$.
We can rewrite this by completing the square:
$2(x^2-4x+4) + 2(y^2-6y+9) = -18 + 8 + 18$
$2(x-2)^2 + 2(y-3)^2 = 8$
$(x-2)^2 + (y-3)^2 = 4$.
When the origin is shifted to $(2,3)$,the equation becomes $X^2 + Y^2 = 4$,where $X = x-2$ and $Y = y-3$.
Rotating the axes by an angle $\theta = 45^{\circ}$ does not change the form of the circle $X^2 + Y^2 = r^2$,as $X^2 + Y^2 = (X' \cos \theta - Y' \sin \theta)^2 + (X' \sin \theta + Y' \cos \theta)^2 = X'^2 + Y'^2$.
Thus,the transformed equation remains $x^2+y^2=4$.

(C) Let $(x, y)$ be the original coordinates and $(X, Y)$ be the new coordinates after rotating the axes 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 $(X, Y) = (2, -6)$ and $\theta = 135^{\circ}$:
$x = 2 \cos 135^{\circ} - (-6) \sin 135^{\circ}$
$y = 2 \sin 135^{\circ} + (-6) \cos 135^{\circ}$
Since $\cos 135^{\circ} = -\frac{1}{\sqrt{2}}$ and $\sin 135^{\circ} = \frac{1}{\sqrt{2}}$:
$x = 2 \left(-\frac{1}{\sqrt{2}}\right) + 6 \left(\frac{1}{\sqrt{2}}\right) = \frac{-2 + 6}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
$y = 2 \left(\frac{1}{\sqrt{2}}\right) - 6 \left(-\frac{1}{\sqrt{2}}\right) = \frac{2 + 6}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4 \sqrt{2}$
Thus,the original coordinates are $(2 \sqrt{2}, 4 \sqrt{2})$.

(B) Step $1$: Shift the origin to $(2,3)$. Substitute $x = X+2$ and $y = Y+3$ into the equation $3x^2+2xy+3y^2-18x-22y+50=0$.
$3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2-18(X+2)-22(Y+3)+50=0$
$3(X^2+4X+4)+2(XY+3X+2Y+6)+3(Y^2+6Y+9)-18X-36-22Y-66+50=0$
$3X^2+12X+12+2XY+6X+4Y+12+3Y^2+18Y+27-18X-36-22Y-66+50=0$
Simplifying,we get $3X^2+2XY+3Y^2-1=0$.
Step $2$: Rotate the axes by $\theta = \frac{\pi}{3}$ counter-clockwise. The transformation equations are:
$X = x' \cos\frac{\pi}{3} - y' \sin\frac{\pi}{3} = \frac{x' - \sqrt{3}y'}{2}$
$Y = x' \sin\frac{\pi}{3} + y' \cos\frac{\pi}{3} = \frac{\sqrt{3}x' + y'}{2}$
Substitute these into $3X^2+2XY+3Y^2-1=0$:
$3(\frac{x'-\sqrt{3}y'}{2})^2 + 2(\frac{x'-\sqrt{3}y'}{2})(\frac{\sqrt{3}x'+y'}{2}) + 3(\frac{\sqrt{3}x'+y'}{2})^2 - 1 = 0$
$\frac{3}{4}(x'^2+3y'^2-2\sqrt{3}x'y') + \frac{2}{4}(\sqrt{3}x'^2+x'y'-3x'y'-\sqrt{3}y'^2) + \frac{3}{4}(3x'^2+y'^2+2\sqrt{3}x'y') - 1 = 0$
Multiplying by $4$:
$3(x'^2+3y'^2-2\sqrt{3}x'y') + 2(\sqrt{3}x'^2-2x'y'-\sqrt{3}y'^2) + 3(3x'^2+y'^2+2\sqrt{3}x'y') - 4 = 0$
$(3+2\sqrt{3}+9)x'^2 + (-6\sqrt{3}-4+6\sqrt{3})x'y' + (9-2\sqrt{3}+3)y'^2 - 4 = 0$
$(12+2\sqrt{3})x'^2 - 4x'y' + (12-2\sqrt{3})y'^2 - 4 = 0$
Dividing by $2$:
$(6+\sqrt{3})x^2 - 2xy + (6-\sqrt{3})y^2 - 2 = 0$.

(B) 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:
$x' = x \cos \frac{\pi}{4} - y \sin \frac{\pi}{4} = \frac{x - y}{\sqrt{2}}$
$y' = x \sin \frac{\pi}{4} + y \cos \frac{\pi}{4} = \frac{x + y}{\sqrt{2}}$
Substituting these into the given equation $3(x')^2 - 6x'y' + 8(y')^2 = 8$:
$3\left(\frac{x - y}{\sqrt{2}}\right)^2 - 6\left(\frac{x - y}{\sqrt{2}}\right)\left(\frac{x + y}{\sqrt{2}}\right) + 8\left(\frac{x + y}{\sqrt{2}}\right)^2 = 8$
$\frac{3}{2}(x^2 + y^2 - 2xy) - \frac{6}{2}(x^2 - y^2) + \frac{8}{2}(x^2 + y^2 + 2xy) = 8$
Multiplying by $2$:
$3(x^2 + y^2 - 2xy) - 6(x^2 - y^2) + 8(x^2 + y^2 + 2xy) = 16$
$(3 - 6 + 8)x^2 + (-6 + 6 + 16)xy + (3 + 6 + 8)y^2 = 16$
$5x^2 + 10xy + 17y^2 = 16$
$5x^2 + 10xy + 17y^2 - 16 = 0$

(D) When the axes are rotated through an angle $\theta = 45^{\circ}$ in the positive direction,the transformation equations are $x = X \cos \theta - Y \sin \theta$ and $y = X \sin \theta + Y \cos \theta$.
Substituting $\theta = 45^{\circ}$,we get $x = \frac{X-Y}{\sqrt{2}}$ and $y = \frac{X+Y}{\sqrt{2}}$.
Given the transformed equation $17x^2 - 16xy + 17y^2 = 225$,we replace $x$ and $y$ with the expressions in terms of $X$ and $Y$:
$17\left(\frac{X-Y}{\sqrt{2}}\right)^2 - 16\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right) + 17\left(\frac{X+Y}{\sqrt{2}}\right)^2 = 225$
$\Rightarrow \frac{17}{2}(X^2 + Y^2 - 2XY) - \frac{16}{2}(X^2 - Y^2) + \frac{17}{2}(X^2 + Y^2 + 2XY) = 225$
$\Rightarrow \frac{17}{2}X^2 + \frac{17}{2}Y^2 - 17XY - 8X^2 + 8Y^2 + \frac{17}{2}X^2 + \frac{17}{2}Y^2 + 17XY = 225$
$\Rightarrow (\frac{17}{2} - 8 + \frac{17}{2})X^2 + (\frac{17}{2} + 8 + \frac{17}{2})Y^2 = 225$
$\Rightarrow (17 - 8)X^2 + (17 + 8)Y^2 = 225$
$\Rightarrow 9X^2 + 25Y^2 = 225$.
Thus,the original equation is $9x^2 + 25y^2 = 225$.

(C) The given equation is $x^2 + 2xy + y^2 - 1 = 0$,which can be written as $(x + y)^2 = 1$.
Let the new coordinates be $(X, Y)$ after shifting the origin to $(1, 1)$ and rotating the axes by $\theta = 45^{\circ}$.
The transformation equations are:
$x = 1 + X \cos 45^{\circ} - Y \sin 45^{\circ} = 1 + \frac{X - Y}{\sqrt{2}}$
$y = 1 + X \sin 45^{\circ} + Y \cos 45^{\circ} = 1 + \frac{X + Y}{\sqrt{2}}$
Substituting these into $(x + y)^2 = 1$:
$(1 + \frac{X - Y}{\sqrt{2}} + 1 + \frac{X + Y}{\sqrt{2}})^2 = 1$
$(2 + \frac{2X}{\sqrt{2}})^2 = 1$
$(2 + X\sqrt{2})^2 = 1$
$4 + 4\sqrt{2}X + 2X^2 = 1$
$2X^2 + 4\sqrt{2}X + 3 = 0$.
Thus,the transformed equation is $2x^2 + 4\sqrt{2}x + 3 = 0$ (replacing $X$ with $x$).