AP EAPCET 21 May 2025 Shift 2 Paper

We have, 3x2+4y2−xy+k=0 is transformed equation after shifting origin to the point ( α,β ) replacing
x⟶x−α and y⟶y−β
‌∴3(x−α)2+4(y−β)2−(x−α)(y−β)+k=0
‌⇒3(x2+α2−2αx)+4(y2+β2−2βy)
‌‌‌−xy+αy+βx−αβ+k=0
‌⇒3x2+4y2−xy+x(−6α+β)+y(−8β+α)
‌‌‌+3α2+4β2−αβ+k=0
Comparing Eq. (i) with given equation,
3x2+4y2−xy−5x−7y+2=0, we get
‌−6α+β=−5
‌α−8β=−7
‌3α2+4β2−αβ+k=2
On solving Eqs. (ii) and (iii), we get α=β=1
By Eqs. (ii)
‌3+4−1+k=2
⇒‌‌k=−4
∴‌‌α+β−k=1+1+4=6.