the common tangent touching hyperbola (x^2)/9 - (y^2)/16=1and (x^2)/16 - (y^2)/9=-1.plz clear my doubt as soon as possible.thans....

consider the equations of  tangents to both the hyperbola in the slope form which is for the first hyperbola y=mx+sqrt(a^2*m^2-b^2) and for the second one it is y=mx +sqrt(a^2+b^2*m^2) where a=3 b=4.. As it is a common tangent the slope is m and for both of them the y intercept has to be equal. Hence on equating we get the value of m which is the slope of the tangent and hence we get the required tangent.