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Find the geometric location of the points of the process
x^(2)+2y^(2)+3z^(2)+xy+2xz+4yz=8 where the tangent plane is parallel to the
xy plane.

Find the geometric location of the points of the process $x^{2}+2y^{2}+3z^{2}+xy+2xz+4yz=8$ where the tangent plane is parallel to the $xy$ plane.

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Pascal's triangle and the Binomial Theorem

Full solution

Q. Find the geometric location of the points of the process

x^{2}+2y^{2}+3z^{2}+xy+2xz+4yz=8

where the tangent plane is parallel to the

xy

plane.

Identify Equation & Condition: Identify the given equation and the condition for the tangent plane. $\newline$ The equation is $x^2 + 2y^2 + 3z^2 + xy + 2xz + 4yz = 8$ . The tangent plane must be parallel to the xy-plane, which implies the normal to the tangent plane has no z-component, i.e., $\frac{\partial f}{\partial z} = 0$ .
Calculate Partial Derivative: Calculate the partial derivative of the equation with respect to z. $\newline$ Differentiating $x^2 + 2y^2 + 3z^2 + xy + 2xz + 4yz$ with respect to z gives $6z + 2x + 4y$ .
Set Equal & Solve for z: Set the partial derivative equal to zero to find z in terms of x and y. $\newline$ $6z + 2x + 4y = 0$ leads to $z = -\frac{1}{3}(2x + 4y)$ .
Substitute z into Equation: Substitute z back into the original equation to find the relationship between x and y. $\newline$ Substituting $z = -\frac{1}{3}(2x + 4y)$ into the original equation, we get: $\newline$ $x^2 + 2y^2 + 3\left(-\frac{1}{3}(2x + 4y)\right)^2 + xy + 2x\left(-\frac{1}{3}(2x + 4y)\right) + 4y\left(-\frac{1}{3}(2x + 4y)\right) = 8$
Simplify Relationship between x and y: Simplify the equation to find a relationship between x and y. $\newline$ Expanding and simplifying, we get: $\newline$ $x^2 + 2y^2 + \frac{4}{3}(x + 2y)^2 - \frac{2}{3}x(2x + 4y) - \frac{4}{3}y(2x + 4y) = 8$ $\newline$ This simplifies to: $\newline$ $x^2 + 2y^2 + \frac{4}{3}(x^2 + 4xy + 4y^2) - \frac{4}{3}x^2 - \frac{8}{3}xy - \frac{16}{3}y^2 = 8$

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