Mix Examples-Rectangular Cartesian Co-ordinates Questions in English

(C) From triangle $OQ_1Q_2$,by applying the cosine rule:
$Q_1Q_2^2 = OQ_1^2 + OQ_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Using the distance formula,$Q_1Q_2^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$,$OQ_1^2 = x_1^2 + y_1^2$,and $OQ_2^2 = x_2^2 + y_2^2$.
Substituting these into the cosine rule:
$(x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$x_1^2 - 2x_1x_2 + x_2^2 + y_1^2 - 2y_1y_2 + y_2^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$-2x_1x_2 - 2y_1y_2 = -2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Dividing by $-2$,we get:
$OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2 = x_1x_2 + y_1y_2$

(C) Given equation: $ax^2-2xy+by^2-2x+4y+1=0$.
Shift origin to $(0, k)$,so $x=X$ and $y=Y+k$.
Substituting these in the equation: $aX^2-2X(Y+k)+b(Y+k)^2-2X+4(Y+k)+1=0$.
Expanding: $aX^2-2XY-2kX+bY^2+bk^2+2bkY-2X+4Y+4k+1=0$.
Grouping terms: $aX^2-2XY+bY^2-2X(k+1)+2Y(bk+2)+bk^2+4k+1=0$.
To remove first degree terms,coefficients of $X$ and $Y$ must be zero: $k+1=0 \Rightarrow k=-1$ and $bk+2=0$ $\Rightarrow -b+2=0$ $\Rightarrow b=2$.
Now,rotate axes by $\theta = \frac{1}{2} \tan^{-1}(2)$,so $\tan(2\theta) = 2$.
The $xy$ term in the rotated equation is removed if $\tan(2\theta) = \frac{2h}{a-b}$,where $h=-1$.
Thus,$\tan(2\theta) = \frac{2(-1)}{a-b} = \frac{-2}{a-b} = \frac{2}{b-a}$.
Given $\tan(2\theta) = 2$,we have $\frac{2}{b-a} = 2 \Rightarrow b-a = 1$.
Since $b=2$,$2-a=1 \Rightarrow a=1$.
Therefore,$a+b = 1+2 = 3$.

(B) Given the equation $4x^2+2xy+y^2-8x-4y-12=0$. To remove the first degree terms,we shift the origin to $P(a, b)$.
Substituting $x=X+a$ and $y=Y+b$ into the equation,we get $4(X+a)^2+2(X+a)(Y+b)+(Y+b)^2-8(X+a)-4(Y+b)-12=0$.
Expanding and collecting terms,the coefficients of $X$ and $Y$ must be zero:
$8a+2b-8=0 \Rightarrow 4a+b=4$
$2a+2b-4=0 \Rightarrow a+b=2$
Solving these,we get $a=2/3$ and $b=4/3$. Thus,$a+b=2$.
Now,the equation becomes $4X^2+2XY+Y^2+C=0$. To remove the $XY$-term,we rotate the axes by an angle $\theta$ such that $\tan 2\theta = \frac{2h}{A-B}$,where the equation is $AX^2+2hXY+BY^2+C=0$.
Here $A=4, B=1, h=1$. So,$\tan 2\theta = \frac{2(1)}{4-1} = \frac{2}{3}$.
Finally,$a+b+3 \tan 2\theta = 2 + 3 \left(\frac{2}{3}\right) = 2+2=4$.