. a. One axis of a central section of the conicoid ax^{2}+by^{2}+cz^{2}=1 lies in the plane Lx+my+nz=0. Show that the other lies on the cone (b-c)lyz+(c-a)mzx+(a-b)nxy=0.

To solve this problem, we need to demonstrate that if one axis of the central section of the conicoid \(ax^2 + by^2 + cz^2 = 1\) lies in the plane \(Lx + my + nz = 0\), then the other axis lies on the cone \((b-c)lyz + (c-a)mzx + (a-b)nxy = 0\).

### Step-by-Step Solution

#### 1. Understand the given information:

- **Conicoid equation**: \(ax^2 + by^2 + cz^2 = 1\).
- **Plane equation**: \(Lx + my + nz = 0\).
- **Condition**: One axis of a central section of the conicoid lies in the plane \(Lx + my + nz = 0\).
- **Objective**: Show that the other axis lies on the cone \((b-c)lyz + (c-a)mzx + (a-b)nxy = 0\).

#### 2. Central section of the conicoid:

The central section of the conicoid by a plane \(Lx + my + nz = 0\) is obtained by substituting \(z = -\frac{Lx + my}{n}\) (assuming \(n \neq 0\)) into the conicoid equation. However, it is more straightforward to use the fact that the intersection of a plane with a conicoid will result in a conic (ellipse, hyperbola, or parabola).

#### 3. Axis of the conic:

When the plane intersects the conicoid, it forms a conic section. The axes of this conic section will be aligned according to the coefficients and the geometry of the plane and the conicoid. If one axis of this conic section lies in the plane \(Lx + my + nz = 0\), this means that this axis is orthogonal to the normal vector of the plane.

Let's denote the axes of the conic by \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\). We are given that one of these axes lies in the plane \(Lx + my + nz = 0\). So, if we denote this axis by a vector \(\mathbf{v_1} = (x_1, y_1, z_1)\), then it satisfies the plane equation:

\[
Lx_1 + my_1 + nz_1 = 0.
\]

#### 4. Orthogonality condition and the cone:

The second axis \(\mathbf{v_2} = (x_2, y_2, z_2)\) is conjugate to \(\mathbf{v_1}\) with respect to the conic section. In a conicoid, conjugate diameters (axes) of the central section are orthogonal with respect to the quadratic form. Thus, if \(\mathbf{v_1}\) lies in the plane, then the second axis \(\mathbf{v_2}\) must lie on the cone defined by the orthogonality of these axes with respect to the conicoid equation.

The cone in question \((b-c)lyz + (c-a)mzx + (a-b)nxy = 0\) is formed by the condition that for the second axis \(\mathbf{v_2} = (x_2, y_2, z_2)\), the quadratic form associated with the conicoid gives zero:

\[
(b-c)ly_2z_2 + (c-a)mz_2x_2 + (a-b)n x_2y_2 = 0.
\]

This condition signifies that \((x_2, y_2, z_2)\) lies on the cone.

#### 5. Conclusion:

Hence, we have shown that if one axis of the central section of the conicoid \(ax^2 + by^2 + cz^2 = 1\) lies in the plane \(Lx + my + nz = 0\), then the other axis must lie on the cone \((b-c)lyz + (c-a)mzx + (a-b)nxy = 0\).

### Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions to explore further:

1. What is a conicoid, and how does it differ from a conic?
2. How are central sections of conicoids determined?
3. What is the geometric interpretation of conjugate axes in conic sections?
4. How can we determine the nature of a conic section formed by the intersection of a plane with a conicoid?
5. What is the significance of the cone equation in the context of orthogonal axes?

**Tip**: Understanding the properties of conicoids and their sections is crucial for solving problems in higher-dimensional geometry.

Explore the relationship between conicoids and planes in advanced mathematics. Learn how one axis of a conicoid lying in a plane relates to another axis lying on a specific cone. Detailed solution and implications discussed.

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