Question

The point to which the axes are to be translated to eliminate $x$ and $y$ terms in the equation $3x^2-4xy-2y^2-3x-2y-1=0$ is

• A. $(\frac{5}{2},3)$
• B. $(-4,\frac{3}{2})$
• C. $(-2,3)$
• D. $(2,3)$

Answer 1

The correct option is A $(\frac{5}{2},3)$

Given equation is $3x^2-4xy-2y^2-3x-2y-1=0$

Let $(x_1,y_1)$ be a point to which the origin is shifted by translation

Let $(X,Y)$ be the new coordinates of the point $(x,y)$

$\therefore$ the equations of the transformation are $x=X+x_1,y=Y+y_1$

Now the transformed equation is

$3{(X+x_1)}^2-4(X+x_1)(Y+y_1)-2{(Y+y_1)}^2-3(X+x_1)-2(Y+y_1)-1=0$

$\Rightarrow 3(X^2+2X{x}_1+{x_1}^2)-4(XY+X{y}_1+Y{x}_1+x_1{y}_1)-2(Y^2+{y_1}^2+2Y{Y}_1)-3X-x_1-2Y-2y_1-1=0$

$\Rightarrow 3X^2+3{x_1}^2+6X{x}_1-4X{y}_1-4x_{1}Y-4x_1{y}_1-2Y^2-2{y_1}^2+4Y{y}_1-3X-3x_1-2Y-2y_1-1=0$

$\Rightarrow (3X^2-4XY-2Y^2)+(3{x_1}^2-2{y_1}^2-4x_1{y}_1-3x_1-2y_1-1)+2X(3x_1-2y_1-\frac{3}{2})+2Y(-2x_1+2y_1-1)=0$

Solving the first degree terms,we have

$3x_1-2y_1=\frac{3}{2}$

$-2x_1+2y_1=1$

Adding the above equations, we get

$3x_1-2y_1-2x_1+2y_1=\frac{3}{2}+1$

$\Rightarrow x_1=\frac{5}{2}$

From equation ,$-2x_1+2y_1=1$

$\Rightarrow 2y_1=1+2x_1=1+2\times \frac{5}{2}=1+5=6$

$\Rightarrow y_1=\frac{6}{2}=3$

$\therefore (x_1,y_1)=(\frac{5}{2},3)$

Hence the point is $(\frac{5}{2},3)$