A common tangent to the conics x2=6y and 2x2−4y2=9 is

Let y=(mx+c) is tangent to x2=6ysolve the above equations ⇒x2=6(mx+c) So, x2−6mx−6c=0 Put D=b2−4ac=0⇒c=−32m2∴ we get y=mx−32m2…….(1) And given hyperbola equation is 2x2−4y2=9⇒x292−y294=1….(2) Since, equation (1) is a tangent of equation (2) then c2=a2m2−b2⇒94m4=92m2−94⇒m4=2m2−1⇒m4−2m2+1=0⇒m2−12=0⇒m=±1 for m=1 , equation of tangent is x−y=32Therefore, the correct answer is (D).