Quadratic Surfaces: Substitute (a,b,c) into z=y^2-x^2

[Fernando Rios]

Homework Statement

Show that if the point (a,b,c) lies on the hyperbolic paraboloid z=y^2-x^2, then the lines with parametric equations x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t both lie enterily on this paraboloid.

Relevant Equations

z=y^2-x^2
x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t

Substitute (a,b,c) into z=y^2-x^2:
c=b^2-a^2

Substitute the parametric equations of L1 into the equation of the hyperbolic paraboloid in order to find points of intersection:
z=y^2-x^2
c+2(b-a)t=(b+t)^2-(a+t)^2=b^2-a^2+(b-a)t
c=b^2-a^2

 

[fresh_42]

Homework Statement:: Show that if the point (a,b,c) lies on the hyperbolic paraboloid z=y^2-x^2, then the lines with parametric equations x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t both lie enterily on this paraboloid.
Relevant Equations:: z=y^2-x^2
x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t

Substitute (a,b,c) into z=y^2-x^2:
c=b^2-a^2

Substitute the parametric equations of L1 into the equation of the hyperbolic paraboloid in order to find points of intersection:
z=y^2-x^2
c+2(b-a)t=(b+t)^2-(a+t)^2=b^2-a^2+(b-a)t
c=b^2-a^2

Looks good, except that you should write it backward.

We need to show that all points $\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}+t\cdot \begin{bmatrix} 1\\ \pm 1\\ \pm 2b -2a\end{bmatrix}$ obey the rule $y^2-x^2=z.$
\begin{align*}
y^2-x^2&=(b\pm t)^2-(a+t)^2=b^2 \pm 2bt +t^2- a^2-2at-t^2\\
&=c + t(\pm 2b-2a)= c \pm 2(b\mp a)t=z
\end{align*}
Of course, you can handle both cases separately if you like that better.

 

[FactChecker]

I think it looks good except that you should state more clearly what some steps are doing.

Homework Statement:: Show that if the point (a,b,c) lies on the hyperbolic paraboloid z=y^2-x^2, then the lines with parametric equations x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t both lie enterily on this paraboloid.
Relevant Equations:: z=y^2-x^2
x=a+t, y=b+t, z=c+2(b-a)t and x=a+t, y=b-t, z=c-2(b+a)t

Substitute (a,b,c) into z=y^2-x^2:
c=b^2-a^2

What are you proving here? That for $t=0, x_0, y_0, z_0$ are on the hyperbolic paraboloid? If so, you should say that. Or are you just stating what it means to say that (a,b,c) is on the hyperbolic paraboloid?
You should say something like: Since (a,b,c) is on the hyperbolic parabola, $c=b^2-a^2$.

Substitute the parametric equations of L1 into the equation of the hyperbolic paraboloid in order to find points of intersection:
z=y^2-x^2
c+2(b-a)t=(b+t)^2-(a+t)^2=b^2-a^2+(b-a)t

Shouldn't this be $b^2-a^2+2(b-a)t$?

c=b^2-a^2

You should state what this shows. (That x, y, z, satisfy the equation of the hyperbolic paraboloid for any value of t.)

 

[Fernando Rios]

Looks good, except that you should write it backward.

We need to show that all points $\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}+t\cdot \begin{bmatrix} 1\\ \pm 1\\ \pm 2b -2a\end{bmatrix}$ obey the rule $y^2-x^2=z.$
\begin{align*}
y^2-x^2&=(b\pm t)^2-(a+t)^2=b^2 \pm 2bt +t^2- a^2-2at-t^2\\
&=c + t(\pm 2b-2a)= c \pm 2(b\mp a)t=z
\end{align*}
Of course, you can handle both cases separately if you like that better.

Thank you for your answer.

 

[Fernando Rios]

I think it looks good except that you should state more clearly what some steps are doing.

What are you proving here? That for $t=0, x_0, y_0, z_0$ are on the hyperbolic paraboloid? If so, you should say that. Or are you just stating what it means to say that (a,b,c) is on the hyperbolic paraboloid?
You should say something like: Since (a,b,c) is on the hyperbolic parabola, $c=b^2-a^2$.

Shouldn't this be $b^2-a^2+2(b-a)t$?

You should state what this shows. (That x, y, z, satisfy the equation of the hyperbolic paraboloid for any value of t.)

Thank you for your response.

 

[pasmith]

Alternative method:
Let f(x,y,z) = y^2 - x^2 - z. Then r(t) = (A_x,A_y,A_z)t + (a, b, c) lies entirely on the paraboloid if and only if f(r(t)) = 0. We know this is the case when t = 0 (because we are given that (a,b,c) is on the surface), so we need <br /> \begin{split}<br /> 0 &amp;= \frac{df}{dt} \\ &amp;= \frac{dr}{dt} \cdot (-2x(t), 2y(t), -1) \\<br /> &amp;= (A_x,A_y,A_z) \cdot (-2(A_xt + a), 2(A_yt + b), -1) \\<br /> &amp;= 2(A_y^2 - A_x^2)t + 2(bA_y - aA_x) - A_z <br /> \end{split} for all t. This requires A_y^2 - A_x^2 = 0 and hence <br /> (A_x,A_y,A_z) = A_x(1, \pm 1, 2(\pm b - a)). Setting A_x = 1 gives the parametrizations specified in the question.