Question

Find the equation of the plane that passes through the point (2,1,−6) and is tangent to the surface
x^2+4y^3+3z^2=116
at (2,1,−6). Write it in the form indicated below.

Equation: 4(x−2)+___(y−___)+___(z−___)=0

Answer 1

$\displaystyle\textsf{First we rearrange the equation of}\\\textsf{the surface into the form}\, f(x, y, z) = 0\\ f(x, y, z) = x^2+4y^3+3z^2 - 116\\ \textsf{In order to find the normal at}\\\textsf{any particular point in vector}\\\textsf{space we use the Del, or gradient operator:}\\ \Delta(x, y, z) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j + \frac{\partial f}{\partial z}k\\ \frac{\partial f}{\partial x} = 2x, \, \frac{\partial f}{\partial y} = 12y^2, \,\frac{\partial f}{\partial z} = 6z\\ \therefore\Delta (x, y, z) = 2x i + 12y^2 j + 6z k\\ \textsf{So for the particular point.}\, (2,1,−6),\,\\\textsf{the normal vectorto the surface}\\\textsf{is given by:}\\ (a, b, c) = \Delta(2,1,−6) = 4i + 12j - 36k\\ \textsf{So the tangent plane to the surface}\\ x^2+4y^3+3z^2 = 116\\ \textsf{has this normal vector and it also}\\\textsf{passes though the point}\, (2, 1, -6). \\\textsf{It will therefore have a}\\\textsf{vector equation of the form:}\\ ax + by + cz = ax_0 + by_0 + cz_0\\ \therefore a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \\ \implies 4(x - 2) + 12(y - 1) - 36(z + 6) = 0$