Question

The surface which intersects the surfaces of the system z(x + y) = c(3z + 1) orthogonally and which passes through the circle x² + y² = 1, z = 1, is given by
1. x² + y² = 2z³ + z² - 2
2. x² - y² = z³ + z + 1
3. x² - y² = z² + 4
4. x² + y² = z³ + z² + 4

Answer 1

Correct Answer - Option 1 : x² + y² = 2z³ + z² - 2

Given:

Surface is f = \(\frac{z(x+y)}{c(3z+1)}\)

Lagrange's subsidiary equation are \(\frac{\partial x}{\partial f}=\frac{\partial y}{\partial f}=\frac{\partial z}{\partial f}\)

⇒ \(\frac{dx}{\frac{z}{c(3z+1)}}=\frac{dy}{\frac{z}{c(3z+1)}}=\frac{dz}{\frac{x+y}{c(3z+1)^2}}\)

⇒ \(\frac{dx}{z}=\frac{dy}{z}=\left(\frac{3z+1}{x+y}\right)dz\)

Taking 1st and 2nd : \(\frac{dx}{z}=\frac{dy}{z}\)

⇒ \(\int dx=\int dy+c_1\)

⇒ x - y = c₁

Again using (i), \(\frac{dx+dy}{2z}=\left(\frac{3z+1}{x+y}\right)dz\)

\(\int (x+y)(dx+dy)=\int(6z^2+2z)dz+c_2\)

⇒ \(\frac{(x+y)^2}{2}=2z^3+z^2+c_2\)

or (x + y)² = 4z³ + 2z² + c₂

x² + y² + 2xy = 4z³ + 2z² + c₂

1 + 2xy = 4(1)³ + 2(1)² + c₂

⇒ 2xy = 5 + c₂

putting the value of 2xy from equation (ii) into equation (iii),

\(1-c^2_1=5+c_2\)

⇒ \(c_1^2+c_2+4=0\)

\((x-y)^2+(x+y)^2-4z^3-2z^2+4=0\)

\(2(x^2+y^2)=2(2z^3+z^2-2)\) or

\(x^2+y^2=2z^3+z^2-2\)