Question

Transforming to parallel axes through a point $(p, q)$, the equation
$2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$ becomes $2x^2 + 3xy + 4y^2 = 1$. Then

• A. $p = -2, q = 3$
• B. $p = 2, q = - 3$
• C. $p = 3, q = - 4$
• D. $p = - 4, q = 3$

Answer 1

Answer: (B)

Given equations are
$2 x^{2}+3 x y+4 y^{2}+x+18 y+25=0 \ldots$ (i)
$2 x^{2}+3 x y+4 y^{2}+1=0 \ldots$(ii)
Let the origin be transferred to $(p, q)$ axes being parallel to the previous axes; then the equation (i) becomes.
$2\left(x^{\prime}+p\right)^{2}+3\left(x^{\prime}+p\right)\left(y^{\prime}+q\right) $
$+4\left(y^{\prime}+q\right)^{2}+\left(x^{\prime}+p\right)+18\left(y^{\prime}+q\right)+25=0 $
$\Rightarrow 2 x^{2}+2 p^{2}+4 x^{\prime} p+3 x^{\prime} y^{\prime}+3 x^{\prime} q+3 p y^{\prime} $
$+3 p q+4 y^{\prime 2}+4 q^{2}+8 y^{\prime} q+x^{\prime}+p$
$+18 y^{\prime}+18 q+25=0$
$\Rightarrow 2 x^{2}+4 y^{\prime 2}+3 x^{\prime} y^{\prime}+(4 p+3 q+1) x^{\prime}$
$1+(3 p+8 q+18) y^{\prime}+2 p^{2}+3 p q $
$+4 q^{2}+p+25=0$
From Eq. (ii) coefficient of $x^{\prime}$ and $y^{\prime}$ must be zero.
$\therefore 4 p+3 q+1=0$ ... (iii)
$3 p+8 q+18=0$ ... (iv)
By solving Eqs. (iii) and (iv), we get
$p=2, q=-3$