Question

The equation of the reflection of the hyperbola $\frac{(x-4{)}^2}{16}-\frac{(y-3{)}^2}{9}=1$ about the line $x+y-2=0$ is

• A. $\frac{(x+3{)}^2}{16}-\frac{(y+4{)}^2}{16}=1$
• B. $\frac{(x+2{)}^2}{16}-\frac{(y+1{)}^2}{9}=-1$
• C. $\frac{(x+1{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$
• D. $\frac{(x+1{)}^2}{16}-\frac{(y+2{)}^2}{9}=-1$

Answer 1

The correct option is C $\frac{(x+1{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$

Any point on the hyperbola is $(4+4\sec \theta ,3+3\tan \theta )$
Reflection of point with respect to the line $x+y-2=0$ is
$\Rightarrow \frac{h-(4+4\sec \theta )}{1}=\frac{k-(3+3\tan \theta )}{1}=-2\frac{(4+4\sec \theta +3+3\tan \theta -2)}{2}$
$\therefore h=-1-3\tan \theta$ and $k=-2-4\sec \theta$
$\Rightarrow \frac{(h+1{)}^2}{3^2}-\frac{(k+2{)}^2}{4^2}=-1$
$\therefore \frac{(x+1{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$