Janet Forbes

Answered

2022-07-10

I want to find the general equation of the two lines of intersection of a one sheet hyperboloid to its tangent plane for the function

$F(x,y,z)={x}^{2}+{y}^{2}-{z}^{2}=1$

at

$({x}_{0},{y}_{0},{z}_{0})$

The equation of the tangent plane is

${x}_{0}x+{y}_{0}y-{z}_{0}z=1$

$F(x,y,z)={x}^{2}+{y}^{2}-{z}^{2}=1$

at

$({x}_{0},{y}_{0},{z}_{0})$

The equation of the tangent plane is

${x}_{0}x+{y}_{0}y-{z}_{0}z=1$

Answer & Explanation

Dalton Lester

Expert

2022-07-11Added 12 answers

Step 1

The 2 families of skew lines ${L}_{a}$ and ${L}_{b}^{\prime}$ generating hyperboloid with one sheet (H) can be retrieved, starting from its equation

$\begin{array}{}\text{(1)}& {x}^{2}+{y}^{2}-{z}^{2}=1\text{}\text{}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}\text{}\text{}(y-z)(y+z)=(1-x)(1+x),\end{array}$

in the following natural way:

Lines $\begin{array}{}\text{(2)}& \text{Lines}\text{}{L}_{a}:\text{}\{\begin{array}{lll}y-z& =& a(1-x)\\ y+z& =& {\displaystyle \frac{1}{a}}(1+x)\end{array}\end{array}$

Lines $\begin{array}{}\text{(3)}& \text{Lines}\text{}{L}_{b}^{\prime}:\text{}\{\begin{array}{lll}y-z& =& b(1+x)\\ y+z& =& {\displaystyle \frac{1}{b}}(1-x)\end{array}\end{array}$

for any non-zero real number a or b.

Indeed: by multiplication of its 2 equations, (2) $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$ (1) ; implication of equations meaning inclusion of corresponding geometric entities ( $\mathrm{\forall}a,{L}_{a}\subset H$ ) as desired. For the same reason, $\mathrm{\forall}b,{L}_{b}^{\prime}\subset H$ .

Therefore, for a given point $({x}_{0},{y}_{0},{z}_{0})$ , you just have to find the values of coefficients a and b, which is straightforward.

Consider the case of a. From the first equation in (2), one gets:

$\begin{array}{}\text{(4)}& a=\frac{{y}_{0}-{z}_{0}}{1-{x}_{0}}=\frac{{y}_{0}\pm \sqrt{{x}_{0}^{2}+{y}_{0}^{2}-1}}{1-{x}_{0}}\end{array}$

which is valid under the condition that ${x}_{0}\ne 1$ . If ${x}_{0}=1$ , get a instead from the second equation in (2).

Do the same for b and plug these expressions into (2), resp. (3).

The 2 families of skew lines ${L}_{a}$ and ${L}_{b}^{\prime}$ generating hyperboloid with one sheet (H) can be retrieved, starting from its equation

$\begin{array}{}\text{(1)}& {x}^{2}+{y}^{2}-{z}^{2}=1\text{}\text{}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}\text{}\text{}(y-z)(y+z)=(1-x)(1+x),\end{array}$

in the following natural way:

Lines $\begin{array}{}\text{(2)}& \text{Lines}\text{}{L}_{a}:\text{}\{\begin{array}{lll}y-z& =& a(1-x)\\ y+z& =& {\displaystyle \frac{1}{a}}(1+x)\end{array}\end{array}$

Lines $\begin{array}{}\text{(3)}& \text{Lines}\text{}{L}_{b}^{\prime}:\text{}\{\begin{array}{lll}y-z& =& b(1+x)\\ y+z& =& {\displaystyle \frac{1}{b}}(1-x)\end{array}\end{array}$

for any non-zero real number a or b.

Indeed: by multiplication of its 2 equations, (2) $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$ (1) ; implication of equations meaning inclusion of corresponding geometric entities ( $\mathrm{\forall}a,{L}_{a}\subset H$ ) as desired. For the same reason, $\mathrm{\forall}b,{L}_{b}^{\prime}\subset H$ .

Therefore, for a given point $({x}_{0},{y}_{0},{z}_{0})$ , you just have to find the values of coefficients a and b, which is straightforward.

Consider the case of a. From the first equation in (2), one gets:

$\begin{array}{}\text{(4)}& a=\frac{{y}_{0}-{z}_{0}}{1-{x}_{0}}=\frac{{y}_{0}\pm \sqrt{{x}_{0}^{2}+{y}_{0}^{2}-1}}{1-{x}_{0}}\end{array}$

which is valid under the condition that ${x}_{0}\ne 1$ . If ${x}_{0}=1$ , get a instead from the second equation in (2).

Do the same for b and plug these expressions into (2), resp. (3).

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