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

The correct option is B $p=2,q=-3$

Given equation,
$2x^2+3xy+4y^2+x+18y+25=0$ ........(i)
Transforming to parallel axes through a point $(p,q)$
$\therefore 2(x+p{)}^2+3(x+p\left)\right(y+q)+4(y+q{)}^2+(x+p)+18(y+q)+25=0$ [replace $x$ by $p+x$ and $y$ by $y+q$ in equation(i)]
$\Rightarrow 2(x^2+p^2+2px)+3(xy+xq+py+pq)+4(y^2+q^2+2yq)+x+p+18y+18q+25=0$
$\Rightarrow 2x^2+4y^2+3xy+4px+3qx+x)+(3py+8qy+18y)+2p^2+4q^2+3pq+p+18q+25=0$
$\Rightarrow 2x^2+4y^2+3xy+(4p+3q+1)x+(3p+8q+18)y+2p^2+4q^2+3pq+p+18q+25=0$
Comparing with equation
$2x^2+4y^2+3xy=1$
we get, $4p+3q+1=0$......(ii)
and $3p+8q+18=0$........(iii)
Solving $\left(ii\right)$ and $\left(ii\right)$, we get values of $p$ and $q$.
From [$3\times$equation(ii)] $-$[equation(iii)$\times 4$]
$\Rightarrow 12p+9q+3-12p-32q-72=0$
$\Rightarrow -23q-69=0$
$\Rightarrow -23q=69$
$\Rightarrow q=\frac{-69}{23}$
$\therefore q=-3$
Substitute $q=-3$ in equation $4p+3q+1=0$, we get
$\Rightarrow 4p+3(-3)+1=0$
$\Rightarrow 4p-9+1=0$
$\Rightarrow 4p-8=0$
$\Rightarrow 4p=8$
$\therefore p=2$
Hence, $p=2,q=-3$