Question

The transformed equation of $3x^2+4xy-7y^2-14=0$ when origin is shifted to the point $(-1,2)$, and then axes being rotated at an angle of ${90}^{\circ }$ is

• A. $21x^2+3y^2+12x+4y+9=0$
• B. $5x^2-6xy+3y^2+24x+4y+13=0$
• C. $4x^2-3xy+3y^2+12x+4y+9=0$
• D. $7x^2-4xy+3y^2+24x-2y+9=0$

Answer 1

The correct option is D $7x^2-4xy+3y^2+24x-2y+9=0$

Let $O^{\prime }$ be $(-1,2)$. Since the axes are rotated about $O^{\prime }$ by an angle ${90}^{\circ }$ in anticlockwise direction, let $(X,Y)$ be the new coordinates with respect to new axes and $(x,y)$ be the coordinates with respect to old axes. Then, we have
$x=X\cos {90}^{\circ }-Y\sin {90}^{\circ }-1=-(Y+1)$
$y=X\sin {90}^{\circ }+Y\cos {90}^{\circ }+2=X+2$
The new equation will be
$3{(-(Y+1\left)\right)}^2+4(-(Y+1\left)\right)(X+2)-7{(X+2)}^2-14=0$

$\Rightarrow 7X^2-4XY+3Y^2+24X-2Y+9=0$
Hence, new equation of curve is $7x^2-4xy+3y^2+24x-2y+9=0$