Question

Find the surface which is orthogonal to the one parameter system z = cxy(x2 + y2) and which passes through the hyperbola x2 - y2 = a2, z = 0 .

13 Feb 2018
Answer :
Word Count : 486
To find the surface orthogonal to the one-parameter system given by the equation \( z = cxy(x^2 + y^2) \) that passes through the hyperbola \( x^2 - y^2 = a^2 \) and \( z = 0 \), we need to solve for the appropriate surface equation. ### Step 1: Determine the vector normal to the given surface The surface is described by \( z = cxy(x^2 + y^2) \). The gradient of the surface at any point \( (x, y, z) \) will give us the normal vector to the surface. To find the gradient, we first calculate the partial derivatives of \( z \) with respect to \( x \) and \( y \): \[ z = cxy(x^2 + y^2) \] #### Partial derivative with respect to \( x \): \[ \frac{\partial z}{\partial x} = c \left[ y(x^2 + y^2) + x \cdot 2x \cdot y \right] = c y(x^2 + y^2 + 2x^2) \] \[ \frac{\partial z}{\partial x} = c y(3x^2 + y^2) \] #### Partial derivative with respect to \( y \): \[ \frac{\partial z}{\partial y} = c \left[ x(x^2 +
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