1

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?

Dominique
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Aarav
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  • The slopes of the tangents to the two curves at $(1,-4)$ are $-2$ and $4$. – geetha290krm Jan 24 '24 at 08:58
  • The conics $y^2-16xz=0$ and $xy+4z^2=0$ in ${\Bbb P}^2$ have projective duals $16XZ-64Y^2=0$ and $16XY+Z^2=0$ in $({\Bbb P}^2)^{\vee}.$ They intersect in $(X:Y:Z)=(\frac14:-\frac14:1),(\frac{\alpha^2}4:\frac{\alpha}4:1),(-\frac{\alpha}4:\frac{\alpha^2}4:1),(1:0:0),$ where $\alpha=\frac{1+i\sqrt3}2$ satisfying $\alpha^3+1=0.$ The $z=1$ affine real common tangent is $\frac14 x-\frac14 y+1=0$ or $y=x+4.$ – Jan-Magnus Økland Jan 26 '24 at 08:03
  • The projective real adds $x=0,$ and the complex projective adds $\frac{\alpha^2}4 x+\frac{\alpha}4 y+z=0$ and $-\frac{\alpha}4 x+ \frac{\alpha^2}4 y +z=0,$ remembering the universal $Xx+Yy+Zz=0,$ where the dual affines are $Z=1$ and $z=1.$ – Jan-Magnus Økland Jan 26 '24 at 08:03

4 Answers4

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 $$ \bbox[5px,border:2px solid #90B030]{y=x+4}\tag5 $$ enter image description here

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

robjohn
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3

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:

enter image description here

Dominique
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  • For your information: this screenshot is made, using GeoGebra. This tool is not capable of displaying/calculating a common tangent: I have just estimated a point on one graph and made a tangent to the other graph from that one point. – Dominique Jan 24 '24 at 10:32
3

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 $y=x+4$

RandomGuy
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1

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 $

Hosam Hajeer
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