Hey Aman.
I think this will work.
Assume a line y = mx + c touches the two given curves.
Hence, when you take one of the curves, and substitute the value of y, you will get a quadratic equation in x which can have only one solution. Because this quadratic equation contains the terms m and c, you can use b^2 - 4ac = 0. This will get you one equation in m and c. Now , take the second curve and repeat the same thing. This gives you another equation in m and c.
Take these two equations and solve them. Depending on the combinations of the curves, it may be very simple or very lengthy.
You may get upto four tangents that way, assuming that the equations you get are quadratic in both m and c.
Mardava
										

										Let the equation of common tangent to y2 = 4ax and x2 = 4 by be y = mx + c.y = mx + c is the tangent to the parabola y2 = 4ax.is tangent to the parabola x2 = 4 by, then it will cut the parabola x2 = 4 by in two coincidental points.mx2 = 4bm2x + 4abmx2 – 4bm2x – 4ab = 0∴ D = (– 4bm2)2 – 4 × m × (– 4ab) = 0Equation of the common tangent isThus, the equation of the common tangent is
										

										Y=x+2Y=mx+2/m is a tangent of y^2=8xY=mx+_2m^1/2 is tangent of xy=-1They are same Som/m^2=1/m=+_m^1/2