Question

When the origin is shifted to the point $\left(\frac{3}{2}, \frac{3}{2}\right)$ by the translation of coordinate axes.then the transformed equation of $32 x^{2}+8 x y+32 y^{2}-108 x-108 y+99=0$ is

• A. $72 X^{2}+56 Y^{2}-63=0$
• B. $X^{2}-14 X Y-7 Y^{2}-2=0$
• C. $32 X^{2}-16 X Y+32 Y^{2}-225=0$
• D. $32 X^{2}+8 X Y+32 Y^{2}-63=0$

Answer 1

Answer: (D)

Substituting $x=X+\frac{3}{2}, y=Y+\frac{3}{2}$ in the equation
$32 x^{2}+8 x y+32 y^{2}-108 x-108 y+99=0$
we get
$32\left(X+\frac{3}{2}\right)^{2}+8\left(X+\frac{3}{2}\right)\left(Y+\frac{3}{2}\right)+32\left(Y+\frac{3}{2}\right)^{2}$
$-108\left(X+\frac{3}{2}\right)-108\left(Y+\frac{3}{2}\right)+99=0$
$\Rightarrow 32\left[X^{2}+3 X+\frac{9}{4}\right]+8\left(\frac{2 X+3}{2}\right)\left(\frac{2 Y+3}{2}\right)$
$+32\left[Y^{2}+3 Y+\frac{9}{4}\right]-108 X-162-108 Y-162+99=0$
$\Rightarrow 32 X^{2}+96 X+72+2(4 X Y+6 X+6 Y+9)+32 Y^{2}$
$+96 Y+72-108 X-162-108 Y-162+99=0$
$\Rightarrow 32 X^{2}+32 Y^{2}+8 X Y-63=0$