Question

Transform the equation $x^2-3xy+11x-12y+36=0$ to parallel axes through the point $(-4,1)$ becomes $a{x}^2+bxy+1=0$ then $b^2-a=$

• A. $\frac{1}{4}$
• B. $\frac{1}{16}$
• C. $\frac{1}{64}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D $\frac{1}{8}$

$x^2-3xy+11x-12y+36=0$

Let $X=(x+4)$ & $Y=(y-1)$

$\therefore$ $x=X-4$, $y=Y+1$

$\therefore$ eqn becomes

${(X-4)}^2-3(X-4)(Y+1)+11(X-4)-12(Y+1)+36=0$

$\Rightarrow X^2+16-8X-3(XY+X-4Y-4)+11X-44-12Y-12+36=0$

$\Rightarrow X^2-8X-3X+11X-3XY+12Y-12Y+12+16-12+36-44=0$

$\Rightarrow X^2-3XY+8=0$

$\therefore$ $\frac{X^2}{8}-\frac{3}{8}\times 4+1=0$

$\therefore$ $a=\frac{1}{8},b=\frac{-3}{8}$ $\therefore$ $b^2-a=\frac{9}{64}-\frac{1}{64}=\frac{8}{64}=\frac{1}{8}$