Question

The new equation of curve

$12x^{2} + 7xy - 12y^{2} - 17x - 31y - 7 = 0$after removing the first degree terms

• 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: (C)

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

Answer 2

Let$\varphi \equiv 12x^{2} + 7xy - 12y^{2} - 17x - 31y - 7 = 0$.....(i)

∴ $\frac{\partial\varphi}{\partial x} \equiv 24x + 7y - 17 = 0$ and $\frac{\partial\varphi}{\partial y} \equiv 7x - 24y - 31 = 0$

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

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

Shift the origin to (1, –1) then replacing $x = X + 1$and $y = Y - 1$in (i), 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., $12X^{2} + 7XY - 12Y^{2} = 0$

Alternative Method : Here $\alpha = 1$ and $\beta = - 1$ and

$g = - 17/2,f = - 31/2,c = - 7$

$\therefore g\alpha + f\beta + c = - \frac{17}{2} \times 1 - \frac{31}{2} \times - 1 - 7 = 0$

$\therefore$ Removed equation is $aX^{2} + 2hXY + bY^{2} + (g\alpha + f\beta + c) = 0$ i.e., $12X^{2} + 7XY - 12Y^{2} + 0 = 0 \Rightarrow$ $12X^{2} + 7XY - 12Y^{2} = 0$.