Question

If \((h,k)\) is the point to which the origin has to be shifted by translation of axes to remove the terms containing \(x\) and \(y\) from the equation \[ 2x^2+3xy+y^2+4x-8y+5=0 \] then \(2h+k=\)

• A. \(0\)
• B. \(1\)
• C. \(20\)
• D. \(16\)

Hint:

For translation of axes problems, first compare with standard second degree equation and use simultaneous equations formed by eliminating linear terms. This method is faster than direct substitution.

Answer 1

The Correct Option is B

Concept: In coordinate geometry, when we shift origin from old coordinates to a new point \((h,k)\), the transformation equations become \[ x=X+h \] \[ y=Y+k \] where \((X,Y)\) are new coordinates. When the problem asks to remove linear terms involving \(x\) and \(y\), we substitute shifted coordinates into the equation and choose values of \(h\) and \(k\) so that coefficients of first degree terms become zero. For a second degree equation of the form \[ ax^2+2hxy+by^2+2gx+2fy+c=0 \] the shift required to eliminate linear terms is obtained by solving equations \[ ah+hk+g=0 \] \[ hh+bk+f=0 \] Thus we convert the equation into standard form and solve the resulting system.

Step 1: Compare given equation with standard second degree form.
The given equation is \[ 2x^2+3xy+y^2+4x-8y+5=0 \] General second degree form is \[ ax^2+2hxy+by^2+2gx+2fy+c=0 \] Comparing terms: Coefficient of \(x^2\) \[ a=2 \] Coefficient of \(xy\) \[ 2h=3 \] Thus \[ h=\frac32 \] Coefficient of \(y^2\) \[ b=1 \] Coefficient of x \[ 2g=4 \] Hence \[ g=2 \] Coefficient of y \[ 2f=-8 \] Thus \[ f=-4 \]

Step 2: Apply equations required to eliminate linear terms.
The required equations are \[ ah+hk+g=0 \] \[ hh+bk+f=0 \] Substitute known values. First equation becomes \[ 2h+\frac32k+2=0 \] Multiply by 2 for simplification \[ 4h+3k+4=0 \] Second equation becomes \[ \frac32h+k-4=0 \] Multiply by 2 \[ 3h+2k-8=0 \] So equations are \[ 4h+3k=-4 \] \[ 3h+2k=8 \]

Step 3: Solve the simultaneous equations.
Multiply second equation by 3 \[ 9h+6k=24 \] Multiply first equation by 2 \[ 8h+6k=-8 \] Subtracting \[ h=32 \] Substitute into second equation \[ 3(32)+2k=8 \] \[ 96+2k=8 \] \[ 2k=-88 \] \[ k=-44 \]

Step 4: Find required quantity \(2h+k\).
We need \[ 2h+k \] Substitute values \[ 2(32)+(-44) \] \[ 64-44 \] \[ =20 \] Thus required value is \[ \boxed{20} \] Hence correct option becomes \[ \boxed{(3)} \]