=1...(ii) Any tangent to the hyperbola eq.(i) y = mx + c where c=±√a2m2−b2...(iii) But this tangent touches the parabola eq.(ii) also ∴
(mx+c)2
a2
−
x2
b2
=1 ⇒ b2(m2x2+c2+2mcx)−a2x2=a2b2 ⇒ (b2m2−a2)x2+2mcb2x+b2(c2−a2)=0 For the tangency, it should have equal roots (2mcb2)2=4(b2m2−a2).b2(c2−a2) ⇒ 4m2c2b4=4b2(b2m2c2−b2m2a2−a2c2+a4) ⇒ c2=a2−b2m2 ⇒ a2m2−b2=a2−b2m2[usingEq.(iii)] ⇒ (a2+b2)m2=a2+b2 ⇒ m2=1⇒m=±1 Hence, the equation of common tangent are y=±x±√(a2−b2)