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