Q15Mathematics II

Question

Q.5. Find the equation of the cone whose vertex is (3,1,2) and base is the circle $2 x^{2} + 2 y^{2} = 1, z = - 1$ .

Answer 1

Given Lagrange's specific auxiliary equations directly from the PDE:

\frac{d x}{z - y} = \frac{d y}{x - z} = \frac{d z}{y - x}

We systematically employ the highly powerful 'Method of Multipliers' to find exact integrable combinations where the resulting denominator strictly vanishes to zero.

First Independent Integral

Let us rigorously choose the multiplier set $(1, 1, 1)$ :

Each Fraction = \frac{1 \cdot d x + 1 \cdot d y + 1 \cdot d z}{1 ( z - y ) + 1 ( x - z ) + 1 ( y - x )}

= \frac{d x + d y + d z}{z - y + x - z + y - x} = \frac{d ( x + y + z )}{0}

Since the exact denominator is strictly zero, the exact numerator absolutely must be zero:

d (x + y + z) = 0 ⟹ x + y + z = C_{1}

Second Independent Integral

Now, let us rigorously choose the completely different multiplier set $(x, y, z)$ :

Each Fraction = \frac{x \cdot d x + y \cdot d y + z \cdot d z}{x ( z - y ) + y ( x - z ) + z ( y - x )}

= \frac{x d x + y d y + z d z}{x z - x y + x y - y z + z y - z x} = \frac{x d x + y d y + z d z}{0}

Again, the denominator is strictly zero:

x d x + y d y + z d z = 0

Integrating this entire expression mathematically yields:

\frac{x ^{2}}{2} + \frac{y ^{2}}{2} + \frac{z ^{2}}{2} = \frac{C _{2}}{2} ⟹ x^{2} + y^{2} + z^{2} = C_{2}

General Solution

The absolute general solution is strictly expressed identically as an arbitrary function precisely connecting the two independent integrals:

f (x + y + z, x^{2} + y^{2} + z^{2}) = 0

Q15Mathematics II

Question

Answer

First Independent Integral

Second Independent Integral

General Solution