A rod of length eight units moves such that its ends A and B always lie on the lines x−y+2=0 and y+2=0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is 9(x2+αy2+βxy+γx+28y)−76=0, then α−β−γ is equal to :
(A)24
(B)23✓ Correct
(C)21
(D)22
Answer: (B)
Solution
Let A=(α,α+2) on x−y+2=0 and B=(β,−2) on y+2=0.
Since P(h,k) divides AB internally in ratio 2:1,
h=32β+α,k=32(−2)+(α+2)=3α−2
So α=3k+2, and 2β=3h−α=3h−3k−2.
Length condition AB=8:
(α−β)2+(α+4)2=64
Substitute α=3k+2 and β=23h−3k−2:
α−β=3k+2−23h−3k−2=23(3k−h+2),α+4=3k+6=3(k+2)
49(3k−h+2)2+9(k+2)2=64
9[(3k−h+2)2+4(k+2)2]=256
Expanding (3k−h+2)2+4(k+2)2=h2+13k2−6hk−4h+28k+20, and replacing (h,k) with (x,y):
9(x2+13y2−6xy−4x+28y)+180=256
9(x2+13y2−6xy−4x+28y)−76=0
Comparing with 9(x2+αy2+βxy+γx+28y)−76=0: α=13,β=−6,γ=−4.
α−β−γ=13+6+4=23
JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option
