bcce42…: “Below a violet ring”

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A plane perpendicular to the $(x,y)$ -plane contains the point $(3,2,2)$ on the paraboloid $36z=4x^2+9y^2$ . The line tangent to the cross-section of the paraboloid created by this plane is parallel to the $(x,y)$ -plane at this point. Find an equation of the plane. $x\cdot \left (\answer {2}\right ) + y \cdot \left (\answer {3}\right ) + z \cdot \left (\answer {0}\right ) = 12$

Let $\vec {n}=\vector {a,b,c}$ be the normal vector for the plane in question.

We know that $c=0$ because the plane is perpendicular to the $(x,y)$ -plane.

The paraboloid $36z=4x^2+9y^2$ can be thought of as the level surface $0 =4x^2+9y^2-36z$ of some new function of several variables: $G(x,y,z) = 4x^2+9y^2-36z$

The gradient vector is normal to level surfaces.

$\grad G = \vector {8x,18y,-36}$

$\grad G(3,2,2) = \vector {24,36,-36}$

Since line tangent to the cross-section of the paraboloid created by this plane is parallel to the $(x,y)$ -plane at $(3,2,2)$ , the tangent vector of the cross-section is parallel to $\vec {n}\cross \grad G(3,2,2)$ .

$\vec {n}\cross \grad (3,2,2) = \vector {-36b,36a,36a-24b}$

Moreover, since this vector is parallel to the $(x,y)$ -plane, $36a-24b = 0.$

So $3a=2b$ .

This means that $\vec {n}$ is parallel to $\vector {a,3a/2,0}$ , which is also parallel to $\vector {2a,3a,0}$ .

Set $\vec {n}= \vector {2,3,0}$ . Now since we know a point on the plane, we can find the plane.