Question

Find a surface satisfyin r+s=0 and touching the elliptic paraboid z=4x2+y2 along its section by the plane.y=2x+1

Answer 1

Find a surface satisfying $r + s = 0$ and touching the elliptic paraboloid $z = 4x^2 + y^2$ along its section by the plane $y = 2x + 1$. **Solution.**

Here the differential equation is $r + s = 0$ which can be written as

$$\frac{\partial p}{\partial x} + \frac{\partial q}{\partial x} = 0 \text{ or } \frac{\partial}{\partial x}(p + q) = 0$$

Integrate with respect to $x$:

$$p + q = f(y)$$

or

$$\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = f(y)$$

Also from $z = 4x^2 + y^2$

$$\frac{\partial z}{\partial x} = p = 8x$$ $$\frac{\partial z}{\partial y} = q = 2y$$

If $r + s = 0$ touches $z = 4x^2 + y^2$ along its section by the plane $y = 2x + 1$ then the values of $p$ and $q$ for an point on this plane should be equal:

$$p + q = 8x + 2y = f(y)$$

We have

$$\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 8x + 2y$$

Integrate with respect to $y$:

$$z + y \frac{\partial z}{\partial x} = 8xy + y^2$$

From $y = 2x + 1$:

$$2x + 1 - y = 0$$

Multiply by 4:

$$8x - 4y + 4 = 0$$

We have

$$z + 4x^2 + y^2 + 8xy + 8x - 4y + 4 = 0$$

Answer:

$$z + 4x^2 + y^2 + 8xy + 8x - 4y + 4 = 0$$