Tangent plane equation question.

  • Thread starter Pavoo
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Homework Statement



Consider a surface ω with equation:

[tex]x^2 + y^2 + 4z^2 = 16[/tex]

Find an equation for the tangent plane to ω at point (a,b,c).

Homework Equations



Tangent plane, 3 variables:

[tex]f_{1}(a,b,c)(x-a) + f_{2}(a,b,c)(y-b) + f_{3}(a,b,c)(z-c)= 0[/tex]

The Attempt at a Solution



I get at the end:

[tex]ax + by + 4cz = a^2 + b^2 + 4c^2[/tex]

The textbook gives me:

[tex]ax + by + 4cz = a^2 + b^2 + 4c^2 = 16[/tex]

Where does the 16 come from?

Comparing to this problem, as an example:

Find an equation of the tangent plane to the sphere [tex]x^2 + y^2 + z^2 = 6 [/tex] at point (1,-1,2). This on is simple. But not the first above.

And is it possible to solve this by expanding to a fourth variable, such as ω?
 

Answers and Replies

  • #2
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The (a, b, c) point must satisfy the equation of the surface, hence the right-hand side equals 16.
 

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