Coordinate geometry after reflection

For Finding reflection curve we will use matrix method . Since reflection of point $(x,y)$ in line $y=(\tan { \theta } )x$ can be calculate by matrix reflection transformation method , which states that :

$\displaystyle{\begin{pmatrix} X \\ Y \end{pmatrix}=\begin{pmatrix} \cos { 2\theta } & \sin { 2\theta } \\ \sin { 2\theta } & -\cos { 2\theta } \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}$

where $(X,Y)$ are new co-ordinates of reflection point $(x,y)$ . Here $\tan { \theta } =2$ .

$\displaystyle{{ \begin{pmatrix} X \\ Y \end{pmatrix}=\begin{pmatrix} \cfrac { -3 }{ 5 } & \cfrac { 4 }{ 5 } \\ \cfrac { 4 }{ 5 } & \cfrac { 3 }{ 5 } \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\\ \begin{pmatrix} X \\ Y \end{pmatrix}=\begin{pmatrix} \cfrac { -3x+4y }{ 5 } \\ \cfrac { 4x+3y }{ 5 } \end{pmatrix}\\ X=\cfrac { -3x+4y }{ 5 } \quad \& \quad Y=\cfrac { 4x+3y }{ 5 } }}$ .

Since in old co-ordinate system : $xy=1$ and since after reflecting old point in line mirror we get reflection point which is also lie on old curve , by using simple optics laws. Hence also $XY=1\quad \\ \Rightarrow \quad \boxed { 12{ x }^{ 2 }-{ 12y }^{ 2 }-7xy+25=0 }$ .

Let a point $(h.k)$ lie on reflected curve. then its reflection about line $y-2x=0$ will lie on $xy=1$ . mirror image of $(h,k)$ w.r.t $y-2x=0$ is $\frac{x-h}{-2} + \frac{y-k}{1} = \frac{-2(k-2h)}{1^{2} + 2^{2}}$ where $x$ and $y$ are reflected coordinates. This gives us $x = \frac{4k-3h}{5}$ and $y= \frac{4h+3k}{5}$ putting the values of $x and y$ in the given curve we get $(\frac{4k-3h}{5})(\frac{4h+3k}{5}) = 1$ which on simplifying we get relation between $h$ and $k$ which is the required locus as $12k^{2} - 12h^{2} - 7kh + 25$ thus answer is $12-12-7+25 =$ $\boxed {18}$

1. You can do it by CO-ORDINATE GEOMETRY
2. (x-X)/a=(y-Y)/b=-2(aX+bY+c)/(a^{2}+b^{2})
3. Where reflection is taken about ax+by+c=0
4. (x,y) lie on given curve.
5. (X,Y) lie on reflected curve.