Given that (α,β) lies on the circle x2+y2=1. ∴α2+β2=1 or it can be rewritten as
1
9
(9α2+4+12α)+β2=1+
1
9
(4+12α) ⇒
1
9
(3α2+2)2+β2=1+
4
9
(1+3α+1)−
4
9
⇒
1
9
(3α+2)2+β2=
5
9
+
4
9
(3α+2) The locus of (3α+2β) is
1
9
x2+y2=
5
9
+
4
9
x or x2−4x+9y2−5=0 On comparing this equation with ax2+2hxy+by2+2gx+2fy+c=0 ⇒a=1,b=9,h=0,g=−2,f=0,c=−5 Now, Δ=abc+2fgh−af2−bg2−ch2 =1×9×(−5)+2(0)−1(0)2−9(−2)2−0 =−45−36=−81≠0 Now, h2−ab=0−9(1)=−9<0 ∴D≠0and h2<ab, Hence, it is an ellipse.