The equation of a common tangent to the curve $y^2=16x$ and $xy= -4$ is ...

Question

First I solved both the curves and the point of intersection is $(1,-4)$ now the equation of tangent is given by $T=0$ i.e $-4y = 8(x+1)$ but this not the correct answer.

What is wrong in this?

Answer 1

This is not the answer to your question, but just a way to show you that the intersection point of both curves has nothing to do with the common tangent:

Answer 2

The common tangent is not usually related to the point of intersection since each curve may have a different tangent at the intersection point they are the same only when the curves "touch" at that point

In general a common tangent is just a line that is tangent to 2 curves at the same of different points

To find this we use the slope form tangents for both curves

$y=mx\pm4\sqrt{m}\ \ $ for this rectangular hyperbola

$y=mx+\frac{4}{m}\ \ $ for the parabola

Solving the equations for getting the same line $m^\frac{3}{2}=\pm1 \ \ $ which allows one solution $m=1$

The common tangent being $y=x+4$

Answer 3

Using homogenized coordiantes, the equation of the parabola is

$ r^T Q_1 r = 0 $

where $ r = [x, y, 1] $ and $Q_1 = \begin{bmatrix} 0 && 0 && -8 \\ 0 && 1 && 0 \\ -8 && 0 && 0 \end{bmatrix} $

Similarly the equation of the hyperbola in homogenized coordinates is

$ r^T Q_2 r = 0 $

where $Q_2 = \begin{bmatrix} 0 && 1 && 0 \\ 1 && 0 && 0 \\ 0 && 0 && 8 \end{bmatrix} $

The tangent line in homogenized coordinates is $ N^T r = 0 $

And since it is common to both conics, then it satisfies

$ N^T Q_1^{-1} N = 0 $

$ N^T Q_2^{-1} N = 0 $

Now,

$ Q_1^{-1} = \dfrac{1}{8} \begin{bmatrix} 0 && 0 && -1 \\ 0 && 8 && 0 \\ -1 && 0 && 0 \end{bmatrix} $

And

$ Q_2^{-1} = \dfrac{1}{8} \begin{bmatrix} 0 && 8 && 0 \\ 8 && 0 && 0 \\ 0 && 0 && 1 \end{bmatrix} $

Therefore, if we let $N = [x, y, 1] $ , then

$ -2 x + 8 y^2 = 0 $, and $ 16 x y + 1 = 0 $

From the first equation, $x = 4 y^2 $. Substitute that into the second equation,

$ 64 y^3 + 1 = 0$

Hence, $ y = - \dfrac{1}{4} $, and it follow from this that $ x= \dfrac{1}{4} $

Hence the normal vector to the line $ N = \frac{1}{4} \ [1 , -1, 4 ] $

And this means that the tangent line is

$ x - y + 4 = 0 $

i.e.

$ y = x + 4 $

Answer 4

Using the point-slope form of a line: $$ y_0'(x-x_0)=y-y_0\tag1 $$ For the curve $y^2=16x$, the slope is $y'=\frac8y$. Letting $y=s$ at the point of tangency, we get $$ \frac8s\left(x-\frac{s^2}{16}\right)=y-s\iff y=\frac8sx+\frac s2\tag2 $$ For the curve $xy=-4$, the slope is $y'=\frac4{x^2}=\frac{y^2}4$. Letting $y=t$ at the point of tangency, we get $$ \frac{t^2}4\left(x+\frac4t\right)=y-t\iff y=\frac{t^2}4x+2t\tag3 $$ Setting $s=\frac{32}{t^2}$ in $(2)$ gives $$ y=\frac{t^2}4x+\frac{16}{t^2}\tag4 $$ $(3)$ and $(4)$ give the same line when $2t=\frac{16}{t^2}$; that is, $t=2$. This gives $$ \bbox[5px,border:2px solid #90B030]{y=x+4}\tag5 $$

One might also consider the line $x=0$ to be a common tangent because it is tangent to $y^2=16x$, and asymptotic to $xy=-4$.