Combining transformations - class-X

$\\ sin(\frac{\pi}{6})=(\frac{b}{1})\\\therefore b= (\frac{1}{2})\\ so new y-coordinate wil be  = (\frac{-1}{2})\\cos(\frac{\pi}{6})=(\frac{a}{1})\\\therefore a=(\frac{\sqrt3}{2})\\\therefore new x-coordinate =(\frac{\sqrt3}{2})$

To remove the first degree terms of a general conic equation, we shift the origin to (h, k) where h and k satisfy partial derivative equations with respect to x and y set to zero. Differentiating 2x^2 + 5xy + 3y^2 + 6x + 7y + 1 = 0 with respect to x gives 4x + 5y + 6 = 0, and with respect to y gives 5x + 6y + 7 = 0. Solving this system gives x = 1 and y = -2.

Rotating axes by 45 degrees involves substituting x = (X - Y)/sqrt(2) and y = (X + Y)/sqrt(2) into the original equation. Expanding 9((X-Y)/sqrt(2))^2 + 16((X+Y)/sqrt(2))^2 = 144 leads to 9(X^2 - 2XY + Y^2)/2 + 16(X^2 + 2XY + Y^2)/2 = 144, which simplifies to 25X^2 + 14XY + 25Y^2 = 288.

We have

$xy-x+2y=6$

Now,

$ xy-x+2y=6 $

$ \Rightarrow xy-x+2y-6=0 $

$ \Rightarrow xy-x+2y-2=4 $

$ \Rightarrow x\left( y-1 \right)+2\left( y-1 \right)=4 $

$ \Rightarrow \left( x+2 \right)\left( y-1 \right)=4 $

This equation is the form of $XY=C$

Then,

$C=4$

Hence, this is the
answer.

When the point $(x,y)$ changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-2 , y=Y-3$
So, the equation transform to 
$(X-2)^2+3(Y-3)^2+4(X-2)+18(Y-3)+30=0$
$\Rightarrow X^2+3Y^2-1=0$
So, the transformed eqn is $X^2+3Y^2-1=0$

Let the point $(x,y)$ on the line changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-1 , y=Y+2$
So, the equation transform to 
$(X-1)^2-(Y+2)^2+2(X-1)+4(Y+2)=0$
$\Rightarrow X^2-Y^2+3=0$
So, the transformed eqn is $X^2-Y^2+3=0$

The radius of the circle is $5$. Despite rotating the reference frame, the origin still remains the same. Hence, the equation of the circle remains unchanged. 
Hence the new equation is:
$ X^2 + Y^2 =25 $

By  rotation  of  axes,
$ x = x _{1} \cos\alpha -y _{1}\sin\alpha $
$ y = x _{1} \sin \alpha + y _{1} \cos \alpha $
$ x \cos  \alpha  + y \sin  \alpha = P$..........(given)
$ \Rightarrow  x _{1} (\cos^2 \alpha + \sin^{2}\alpha ) + y (\sin  \alpha  \cos  \alpha - \cos  \alpha  \sin  \alpha ) = P$..............(substitute the values of x and y) 
$x _{1} = P$ => X=P.

When axes are rotated through an angle $\theta $, then 

$x=X \cos\theta -Y \sin\theta$

$y=Y \cos\theta -X \sin\theta$

where(x,y) are initial coordinates and (X,Y) are new coordinates

$\therefore x=4\left(\dfrac{-1}{\sqrt{2}}\right)-\left(-3\right)\left(\dfrac{1}{\sqrt{2}}\right)=\dfrac{-1}{\sqrt{2}}$

$y=4\left(\dfrac{1}{\sqrt{2}}\right)-3\left(\dfrac{-1}{\sqrt{2}}\right)=\dfrac{7}{\sqrt{2}}$

$\therefore (x,y)=\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

When axes are rotated through an angle $\theta $, then

$x=\alpha +x^{1}\cos\theta -y^{1}\sin\theta  ; y=\beta +x^{1}\sin\theta +y^{1}\cos\theta $

$\Rightarrow 4=3+\dfrac{x^{1}\sqrt{3}}{2}-\dfrac{y^{1}}{2} ;3=1+\dfrac{x^{1}}{2}+\dfrac{y^{1}\sqrt{3}}{2}$

$\Rightarrow x^{1}\sqrt{3}-y^{1}=2$

$ x^{1}+y^{1}\sqrt{3}=4$

$\Rightarrow x=\dfrac{\sqrt{3}+2}{2}$

$y=\dfrac{2\sqrt{3}-1}{2}$

Shifting the origin to (-2, -3) and rotating by 45 degrees involves two steps: substitution of x = X' - 2, y = Y' - 3, followed by rotation. The resulting equation matches option A.

Since the point $A(2, 1)$ is translated parallel to $x-y =3$, therefore $AA'$ has the same slope as that of $x-y =3$. Therefore, $AA'$ passes through $(2, 1)$ and has the slope of $1$. 

$x=r\cos \alpha=4$

Given, points are $A=(1,0)$ and $B=(2,1)$

Since, the axes are rotated through an angle $45$, then we replace $\left( x,y \right) $ by
$\left( x\cos { 45 } -y\sin { 45 } ,x\sin { 45 } +y\cos { 45 }  \right) $
Given equation is $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$
Therefore, $ 3{ \left( \dfrac { x }{ \sqrt { 2 }  } -\dfrac { y }{ \sqrt { 2 }  }  \right)  }^{ 2 }+3\dfrac { x+y }{ \sqrt { 2 }  } +2\dfrac { x-y }{ \sqrt { 2 }  } \dfrac { x+y }{ \sqrt { 2 }  } =2$
$\Rightarrow \dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }+2xy \right) +\dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }-2xy \right) +\dfrac { 2 }{ 2 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$
$\Rightarrow  4{ x }^{ 2 }=2{ y }^{ 2 }=2$
$\Rightarrow  2{ x }^{ 2 }+{ y }^{ 2 }=1$