Question

The no of common tangents to y^2=2012 x and xy=(2013)^2 is

Answer 1

The given curves are

1.$y^2=2012 x$  Which is a parabola having vertex at the origin.

2. $xy=(2013)^{2}$ is the equation of rectangular hyperbola having center at the origin. the curve will never touch either positive or negative side of X and Y axis.

Parabola and Rectangular Hyperbola will intersect at a point which lies in first quadrant.

Drawn the graph for you.

y'=1006/y, and for second curve y'=-(2013)²/x²

→1006/y = -(2013)²/x²....(1)

Putting y=(2013)²/x in 1, we get

x=$-\frac{2013^\frac{4}{3}}{1006\frac{1}{3}}$

We get negative value of y. So (x,y) lies in third quadrant.

As , y²=2012 x, so putting the value y we get a positive value of x which lies in fourth quadrant.So one point which lies on rectangular hyperbola is in third quadrant which is (-p,-q) and another is (r,-s) which lies on parabola.

so there is one tangent passing through these two points.

As you can see there is a common tangent between two curves.

So, number of common tangents=1