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.

Question

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.

Bytelearn · Accepted Answer

Identify Equation & Condition:  Identify the given equation and the condition for the tangent plane. 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. 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. $6z + 2x + 4y = 0$ 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. Substituting $z = -\frac{1}{3}(2x + 4y)$ into the original equation, we get: $$x^2 + 2y^2 + 3\left(-\frac{1}{3}(2x + 4y)\right)^2 + xy + 2x\left(-\frac{1}{3}(2x + 4y)\right) + 4y\left(-\frac{1}{3}(2x + 4y)\right) = 8$$Simplify Relationship between x and y:  Simplify the equation to find a relationship between x and y. Expanding and simplifying, we get: $$x^2 + 2y^2 + \frac{4}{3}(x + 2y)^2 - \frac{2}{3}x(2x + 4y) - \frac{4}{3}y(2x + 4y) = 8$$ This simplifies to: $$x^2 + 2y^2 + \frac{4}{3}(x^2 + 4xy + 4y^2) - \frac{4}{3}x^2 - \frac{8}{3}xy - \frac{16}{3}y^2 = 8$$

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.

Full solution

More problems from Pascal's triangle and the Binomial Theorem

Related topics

Find the geometric location of the points of the process x2+2y2+3z2+xy+2xz+4yz=8x^{2}+2y^{2}+3z^{2}+xy+2xz+4yz=8x2+2y2+3z2+xy+2xz+4yz=8 where the tangent plane is parallel to the xyxyxy 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.