Tangent Plane Problem: Where am I wrong?

[gikiian]

1. Homework Statement :

Find the points on the hyperboloid of two sheets with equation x²-2y²-4z²=16 at which, the tangent plane is parallel to the plane 4x-2y+4z=5.

Homework Equations

The hyperboloid with two sheets: x²-2y²-4z²=16

The given plane: 4x-2y+4z=5

Equation of tangent plane: Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0 ;
Fx, Fy and Fz are partial derivatives
3. The Attempt at a Solution :

For the hyperboloid, we can say that

F(x,y,z)=> x² - 2y² - 4z² - 16= 0​

Equation of the tangent plane is given by:

Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0​

so we need to find Fx, Fy and Fz.

Fx = 2x
Fy = -4y
Fz = -8z​

At P(x₀,y₀,z₀) on the hyperboloid, the partial derivatives will be:

Fx = 2x₀
Fy = -4y₀
Fz = -8z₀​

Putting these results in the equation of the tangent plane, we have:

2x₀(x-x₀) - 4y₀(y-y₀) – 8z₀(z-z₀) = 0
(2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀²​

Since this plane is parallel to the given plane 4x - 2y + 4z = 5, we multiply one of the planes by any constant, say λ, to get the other one. I chose to multiply λ with the equation of the plane tangent to the hyperbpoloid:

λ [ (2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀² ]​

Next, by comparing the respective co-efficients of x, y and z of the given and the obtained equation, we get:

2x₀ λ = 4 ------------(i)
4y₀λ = 2-------------(ii)
-8z₀λ = 4 ------------ (iii)
( 2x₀²-4y₀²-8z₀² ) λ = 5 ------------- (iv)

From (i), x₀ = 2 / λ -------------(iv)
From (ii), y₀ = 1 / 2λ ---------(v)
From (iii), z₀ = - 1 / 2λ ----------(vi)​

Putting these values in equation (iv), we get:

( 2(2/λ)²-4(1/2λ)²-8(-1/2λ)² ) λ = 5
( 8/ λ² - 1/ λ² - 2/ λ² ) λ = 5
( 5/λ² ) λ = 5
λ² - λ = 0
λ(λ-1 )= 0
λ = 0……… or ……… λ = 1​

Values of x₀, y₀ and z₀ are not possible for λ = 0, hence for λ = 1, we have from eq. (iv), (v) and (vi):

x₀ = 2
y₀ = 1/2
z₀ = -1/2​

Hence, the point on the hyperbola having the tangent plane parallel to the plane is ( 2 , 1/2 , -1/2 )

My comments:

My solution has something wrong in it. Also, this point does not satisfy the equation of the hyperboloid.

The real answers are:
( 8(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) )
and ( -8(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) )

Please help me Identify my mistake. Thanks.

 

[Stephen Tashi]

Fx = 2x
Fy = -4y
Fz = -8z​

Why did you get Fz = -8z ?

 

[HallsofIvy]

1. Homework Statement :

Find the points on the hyperboloid of two sheets with equation x²-2y²+4z²=16 at which, the tangent plane is parallel to the plane 4x-2y+4z=5.

Homework Equations

The hyperboloid with two sheets: x²-2y²+4z²=16

The given plane: 4x-2y+4z=5

Equation of tangent plane: Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0 ;
Fx, Fy and Fz are partial derivatives



3. The Attempt at a Solution :

For the hyperboloid, we can say that

F(x,y,z)=> x² - 2y² + 4z² - 16= 0​

Equation of the tangent plane is given by:

Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0​

so we need to find Fx, Fy and Fz.

Fx = 2x
Fy = -4y
Fz = -8z​

You have the sign wrong here. $F_z= 8z$

At P(x₀,y₀,z₀) on the hyperboloid, the partial derivatives will be:

Fx = 2x₀
Fy = -4y₀
Fz = -8z₀​

Again, sign wrong.

Putting these results in the equation of the tangent plane, we have:

2x₀(x-x₀) - 4y₀(y-y₀) – 8z₀(z-z₀) = 0
(2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀²​

Since this plane is parallel to the given plane 4x - 2y + 4z = 5, we multiply one of the planes by any constant, say λ, to get the other one. I chose to multiply λ with the equation of the plane tangent to the hyperbpoloid:

λ [ (2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀² ]​

Next, by comparing the respective co-efficients of x, y and z of the given and the obtained equation, we get:

2x₀ λ = 4 ------------(i)
4y₀λ = 2-------------(ii)
-8z₀λ = 4 ------------ (iii)​

Of course, that should be $8z_0\lambda= 4$. You could have got these more simply by noting that the normal to the surface, $\nabla x^2- 2y^2+ 16z^2= 2x\vec{i}- 4y\vec{j}+ 8z\vec{k}$ must be parallel to the normal to the plane, $4\vec{i}- 2\vec{j}+ 4\vec{k}$.

( 2x₀²-4y₀²-8z₀² ) λ = 5 ------------- (iv)

Now, this is a major error- and the reason why your answer is not even on the hyperbola. You had the $\lambda$ in the first three equations because the two planes only had to be parallel, not the same plane. But the point must be on[/b[] the hyperbola so you must have
$2x_0^2- 4y_0^2+ 8z_0^2=5$- there is no "$\lambda$" here.
(and note the +8, not -8). Finally, it should be equal to 16, not 5!

From (i), x₀ = 2 / λ -------------(iv)
From (ii), y₀ = 1 / 2λ ---------(v)
From (iii), z₀ = - 1 / 2λ ----------(vi)

​

$z_0= 1/2\lambda$

Putting these values in equation (iv), we get:

( 2(2/λ)²-4(1/2λ)²-8(-1/2λ)² ) λ = 5
( 8/ λ² - 1/ λ² - 2/ λ² ) λ = 5
( 5/λ² ) λ = 5
λ² - λ = 0​

Again, that last $\lambda$ should not be there- you have only $5/\lambda^2= 5$. And, it is equal to 16, not 5.
But even if that were correct, you have an algebra error here- $(5/\lambda^2)\lambda= 5/\lambda= 5$ gives $1/\lambda= 1$- there is no subtraction.

[λ(λ-1 )= 0
λ = 0……… or ……… λ = 1

$5/\lambda^2= 16$ gives $\lambda^2= 5/16$ so that $\lambda= \pm \sqrt{5}/4$

Values of x₀, y₀ and z₀ are not possible for λ = 0, hence for λ = 1, we have from eq. (iv), (v) and (vi):

x₀ = 2
y₀ = 1/2
z₀ = -1/2​

Hence, the point on the hyperbola having the tangent plane parallel to the plane is ( 2 , 1/2 , -1/2 )

The equations should be $2x_0\lambda= 4$, $4y_0\lambda= 2$, and $8z_0\lambda= 4$. Solve those with $\lambda= \sqrt{5}/4$ and $\lambda= -\sqrt{5}/4$.

My comments:

My solution has something wrong in it. Also, this point does not satisfy the equation of the hyperboloid.

The real answers are:
( 8(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) )
and ( -8(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) )

Please help me Identify my mistake. Thanks.

 

[gikiian]

@HallsofIvy & Stephen Tahsi:

Sir I am terribly sorry; the equation meant to be a hyperboloid in two sheets, and it it as follows:

x²-2y²-4z²=16​

So Fz=-8z.

The site which you called as the "major error" is basically a concern.

Thanks for tackling :)

P.S. I corrected the problem question above.