The correct option is D x2=2(cost+3),y2=2(cos2t2−1)
Option (a),
txa−yb+t=0⋯(i)
xa+tyb−1=0⋯(ii)
⇒t=by(1−xa)
Put this value in (i), we get
by(1−xa)(1+xa)−yb=0
⇒by(1−x2a2)−yb=0
x2a2+y2 b2=1
which represents an ellipse.
Option (b)
x=a2(t+1t)
⇒2xa=t+1t⋯(iii)
y=b2(t−1t)
⇒2yb=t−1t⋯(iv)
⇒4x2a2−4y2 b2=4
⇒x2a2−y2b2=1
which represents a hyperbola
Option (c)
x=et+e−t⋯(v)
y=et−e−t⋯(vi)
⇒x2−y2=4 which is a hyperbola
Option (d)
x2=2(cost+3)⋯(vii)
⇒x2=4cos2t2+4
⇒x24=cos2t2+1
y2=2(cos2t2−1)⋯(viii)
⇒y22=cos2t2−1
⇒x28−y24=1
which represents a hyperbola.