Animated Solution for Mathematics - Conic Sections: The equation of a common tangent to the curves, $y^2=16x$ and $xy=-4$ is :

We are tasked with finding the common tangent to the parabola $y^2=16x$ and the rectangular hyperbola $xy=-4$.

The parabola $y^2=16x$ is a wide, sweeping arc opening to the right. The hyperbola $xy=-4$ is a sharp, dual-branched curve residing in the second and fourth quadrants.

Every great journey begins with a tool. For our parabola $y^2=16x$, we recall the standard form $y^2=4ax$.

By comparing the two, we see that $4a=16$, which gives us $a=4$.

We invoke the standard equation of a tangent to a parabola with slope $m$:

Substituting our value of $a$, we obtain the candidate line:

This line is guaranteed to be tangent to the parabola for any non-zero slope $m$.

To force this line to also touch the hyperbola $xy=-4$, we substitute the tangent line equation into the hyperbola's equation:

Expanding this expression, we get:

Bringing all terms to one side, we arrive at the quadratic equation:

For the line to be tangent to the hyperbola, it must intersect at exactly one point. In the language of quadratics, this means the equation must have equal roots, requiring the discriminant $D=B^2-4AC$ to be zero.

Here, our coefficients are $A=m$, $B=\frac{4}{m}$, and $C=4$. Plugging these into the condition $D=0$:

This simplifies to:

Dividing by 16 and rearranging, we find:

The only real solution for the slope is $m=1$.

We substitute $m=1$ back into our original tangent equation $y=mx+\frac{4}{m}$:

This simplifies to $y=x+4$. Rearranging into the standard general form, we obtain the final equation:

This is the perfect, singular line that bridges the parabola and the hyperbola. You have successfully navigated the geometry and mastered the algebra.