Question

The new equation of the curve $12x^2+7xy-12y^2-17x-31y-7=04$ after removing the first degree terms is

• A. $12X^2-7XY-12Y^2=0$
• B. $12X^2+7XY-12Y^2=0$
• C. $12X^2+7XY+12Y^2=0$
• D. none of these

Answer 1

Answer: (B)

$12X^2+7XY-12Y^2=0$

Let

$\varphi \equiv 12x^2+7xy-12xy-12y^2-17x-31y-7=0$

$\therefore \frac{\partial \varphi }{\partial x}=24x+7y-17=0$

and, $\frac{\partial \varphi }{\partial y}=7x-24y-31=0.$

Their point of intersection is $(x,y)\equiv (1.-1).$

Here, $\alpha =1,\beta =-1$

Shifting the origin to $(1,-1),$ by replacing

$x=X+1$ and $y=Y-1$ in (1), the required equation is

$12{(X+1)}^2+7(X+1)(Y-1)-12{(Y-1)}^2-17(X+1)-31(Y-1)-7=0$

i.e $12X^2+7XY-12Y^2=0$