If the tangents drawn to the hyperbola 4y² = x² + 1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is

4y2=x2+1⇒–x2+4y2=1⇒−x212+y2(12)2=1a=1,b=12Let, tangent to the curve is at point(x1,y1).∴4×2y.dydx=2x⇒dydx=2x18y1=x14y1∴Eqnof tangent:y=mx+c⇒y=x14y1⋅x+c⇒y1=x1y14y1+c⇒c=y1=x214y1=4y21–x214y1=14y1⇒y=x14y1x+14y1⇒4y1y=x1x+1…. (I)Intersectsxaxis at(𕒵x1,0)Andyaxis at=(0,14y1)h=󔼔x1x1=󔼔hy1=18kMidpoint :(󔼔x1,18y1):(h,k)4y21≠x21+1⇒4(18k)2=(󔼔h)2+1⇒116k2=14h2+1⇒1=16k24h2+16k2h2=4k2+16h2k.x2–4y2–16x2y2=0This is the required equation.