Calculate the slope of the tangent line to the intersection curve of the surface Z= ½√24-X² -2Y² with the plane Y=2 at the point (2,2, √3)

First, we need to find the partial derivatives of the surface:
Z = \frac{1}{2} \sqrt{24 - X^2 - 2Y^2}

The partial derivatives with respect to \(X\) and \(Y\) are:
\frac{\partial Z}{\partial X} = \frac{\partial}{\partial X} \left( \frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right)
\frac{\partial Z}{\partial Y} = \frac{\partial}{\partial Y} \left( \frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right)

Calculate these partial derivatives:
\frac{\partial Z}{\partial X} = \frac{\partial}{\partial X} \left( \frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right) = \frac{1}{2} \cdot \frac{1}{2 \sqrt{24 - X^2 - 2Y^2}} \cdot (-2X) = -\frac{X}{2\sqrt{24 - X^2 - 2Y^2}}
\frac{\partial Z}{\partial Y} = \frac{\partial}{\partial Y} \left( \frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right) = \frac{1}{2} \cdot \frac{1}{2 \sqrt{24 - X^2 - 2Y^2}} \cdot (-4Y) = -\frac{2Y}{2\sqrt{24 - X^2 - 2Y^2}} = -\frac{Y}{\sqrt{24 - X^2 - 2Y^2}}

Evaluate the partial derivatives at the given point \((2, 2, \sqrt{3})\):

Partial derivative with respect to \(X\):
\frac{\partial Z}{\partial X}(2, 2, \sqrt{3}) = -\frac{2}{2\sqrt{24 - 2^2 - 2(2)^2}} = -\frac{2}{2\sqrt{24 - 4 - 8}} = -\frac{2}{2\sqrt{12}} = -\frac{2}{2(2\sqrt{3})} = -\frac{1}{2\sqrt{3}} = -\frac{\sqrt{3}}{6}

Partial derivative with respect to \(y\):
\frac{\partial Z}{\partial Y}(2, 2, \sqrt{3}) = -\frac{2}{\sqrt{24 - 2^2 - 2(2)^2}} = -\frac{2}{\sqrt{24 - 4 - 8}} = -\frac{2}{\sqrt{12}} = -\frac{2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} = \frac{-1}{\sqrt{3}} = -\sqrt{3}

The slope of the tangent plane at \((2,2,\sqrt{3})\) in the direction of \(Y\) is given by:
m = -\frac{\partial Z}{\partial Y} = -\sqrt{3}

Therefore, the slope of the tangent line in the direction of \(Y\) when \(Y = 2\):
m = -2