Prove that ax^2+by^2+cz^2 +2ux+2vy+2wz+d=0 represent a cone if u^2/a+v^/b+w^2/c=d

To demonstrate that the equation \( ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz + d = 0 \) represents a cone under the condition \( \frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c} = d \), we need to analyze the structure of the equation and the implications of the given condition.

Understanding the Equation

The general form of the equation can be rearranged to highlight its quadratic nature. We can express it as:

\( ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz = -d \)

This is a quadratic equation in three variables \( x, y, z \). The coefficients \( a, b, c \) are associated with the squared terms, while \( u, v, w \) are the coefficients of the linear terms.

Identifying the Cone Structure

A cone in three-dimensional space can be represented by a quadratic equation that has a specific relationship between its coefficients. The condition for a cone is that the quadratic form must be degenerate, meaning it can be factored into linear components.

To analyze this, we can complete the square for each variable:

• For \( x \): \( ax^2 + 2ux = a\left(x^2 + \frac{2u}{a}x\right) = a\left(x + \frac{u}{a}\right)^2 - \frac{u^2}{a} \)
• For \( y \): \( by^2 + 2vy = b\left(y^2 + \frac{2v}{b}y\right) = b\left(y + \frac{v}{b}\right)^2 - \frac{v^2}{b} \)
• For \( z \): \( cz^2 + 2wz = c\left(z^2 + \frac{2w}{c}z\right) = c\left(z + \frac{w}{c}\right)^2 - \frac{w^2}{c} \)

Substituting these completed squares back into the equation gives:

\( a\left(x + \frac{u}{a}\right)^2 + b\left(y + \frac{v}{b}\right)^2 + c\left(z + \frac{w}{c}\right)^2 - \left(\frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c}\right) = -d \)

Applying the Given Condition

Now, substituting the condition \( \frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c} = d \) into our equation results in:

\( a\left(x + \frac{u}{a}\right)^2 + b\left(y + \frac{v}{b}\right)^2 + c\left(z + \frac{w}{c}\right)^2 - d = -d \)

This simplifies to:

\( a\left(x + \frac{u}{a}\right)^2 + b\left(y + \frac{v}{b}\right)^2 + c\left(z + \frac{w}{c}\right)^2 = 0 \)

Interpreting the Result

The equation \( a\left(x + \frac{u}{a}\right)^2 + b\left(y + \frac{v}{b}\right)^2 + c\left(z + \frac{w}{c}\right)^2 = 0 \) indicates that each squared term must equal zero since \( a, b, c \) are positive constants. This leads to:

• \( x + \frac{u}{a} = 0 \)
• \( y + \frac{v}{b} = 0 \)
• \( z + \frac{w}{c} = 0 \)

These equations describe a line through the point \( \left(-\frac{u}{a}, -\frac{v}{b}, -\frac{w}{c}\right) \) in three-dimensional space. The set of all such lines forms a cone, with the vertex at this point and extending infinitely in both directions along the line.

Conclusion

Thus, we have shown that the original equation represents a cone when the condition \( \frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c} = d \) holds true. The relationship between the coefficients ensures that the quadratic form is degenerate, confirming the conical structure of the solution set.