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This is the work I have so far:

$f_x = 2x_0,\ \ \ x_0 = x_0 \ \ \ \ \ \ f_x = 2x_0$

$f_y = 6y_0, \ \ \ \ \ y_0 = 0 \ \ \ \ \ \ f_y = 0$

$f_z = 20z_0, \ \ \ \ \ z_0 = 0 \ \ \ \ \ \ f_z = 0 \\$

$f_x(x-x_0) + f_y(y-y_0) + f_z(z-z_0) = 0$

$2x_0(1 - x_0) = 0$

$2x_0 - 2x_0^2$

$x = 1$

Apparently, the answer is actually e) but I am unsure of what I am doing wrong.

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  • $\begingroup$ evaluate the derivatives at $ (1, -1, 1 )$ not $( x_0, 0, 0)$ $\endgroup$
    – WW1
    May 9 '15 at 20:10
  • $\begingroup$ Why would that be so? In my textbook the equation of a tangent plane is written as $f_x(P_0)(x - x_0) + f_y(P_0)(y - y_0) + f_z(P_0)(z - z_0) = 0$. I understand that switching it to (1, -1, 1) will give the right answer, I just don't understand why if the formula given in my textbook calls for $(x_0, 0, 0)$ $\endgroup$
    – Robert
    May 9 '15 at 20:12
  • $\begingroup$ In your textbook's formula, $x_0$ is the $x$-coordinate of the point on the surface given by $f(x,y,z)=0$, and so on for $y_0$ and $z_0$. You are trying to make those variables to be the intersection of the tangent plane with the $x$-axis, quite a different thing. You need to first find the equation of the tangent plane, using the book's formula with $x_0=1,y_0=-1,z_0=1$, and then solve that equation for $x$ after substituting $y=0,z=0$. $\endgroup$ May 9 '15 at 20:24
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It is unfortunate that the question named the x-intercept $x_0$ which appears as a symbol in your formula making reference to the point on the ellipsoid. The question might be clearer if they said the x-intercept was at $(a,0,0)$ and asked you to find $a$

you should use $P_0=(x_0, y_0, z_0)=(1,-1,1)$

so the equation of the tangent plane is $2(x-1) -6(y+1)+20(z-1)=0$ and the x-intercept is at $(14,0,0)$

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  • $\begingroup$ Quick question: shouldn't the 3(z-1) actually be 20(z-1)? $\endgroup$
    – Robert
    May 9 '15 at 21:25
  • $\begingroup$ I got the equation $2(x−1)−6(y+1)+20(z−1)=0. \\ f_x = 2x = 2(1) = 2. \\ f_z = 20z = 20(1) = 2$ $\endgroup$
    – Robert
    May 9 '15 at 21:32
  • $\begingroup$ you are correct about the $20(z-1)$ I have edited my answer accordingly $\endgroup$
    – WW1
    May 9 '15 at 21:35
  • $\begingroup$ For Future Viewers: I made an error in the original post. The answer was actually e) and the correct equation is 2(x−1)−6(y+1)+20(z−1)=0. $\endgroup$
    – Robert
    May 9 '15 at 21:51

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