The equation of the common tangent to the curves y2=8x and xy=−1 is

A point on the curve xy=−1 is of the form (t,−1/t). Now y=−1x⇒dydx=1x2⇒dydx(t,−1/t)=1t2∴ Equation of tangent to the curve xy=−1 at (t,−1/t) is y+1t=1t2(x−t)⇒y=1t2x−2t For this line to be tangent to the parabola y2=8x it  should be of the form y=mx+2m so m=1t2 and −2t=2m or m=−t∴ −t=1t2⇒−t3=1⇒t=−1 Thus the required tangent  is _y=x+2.