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 $p=2,q=-3$

$2x^2+3xy+4y^2+x+18y+25=0$ ......... $\left(i\right)$

After transforming equation $\left(i\right)$ through $(p,q)$ it becomes

$2x^2+3xy+4y^2=1$ ...... $\left(ii\right)$

Then, the point $(p,q)$ satisfies $\left(i\right)$ and $\left(ii\right)$

Differentiate $\left(i\right)$ w.r.t $x$ and $y$ respectively, we get

$4x+3y+1=0$ ....... $\left(iii\right)$ and

$3x+8y+18=0$ ....... $\left(iv\right)$

$\because (p,q)$ satisfies $\left(i\right)$ and $\left(ii\right)$

$⟹$ Point $(p,q)$ satisfies $\left(iii\right)$ and $\left(iv\right)$

So, we get

$4p+3q+1=0$ ........ $\left(v\right)$
$3p+8q+18=0$ ....... $\left(vi\right)$

Solving $\left(v\right)$ and $\left(vi\right)$ simultaneously we get
$\therefore p=2,q=-3$.